Calculus Volume 3 · Chapter 4 · Section 4.5
The Chain Rule
When z depends on x and y, and they both depend on t, z changes along every route at once. Add the routes.
Colour key
key idea
· round handles can be dragged · page links open the textbook
Level 1 · see it A rate along a path before any formula
How fast is the fly warming up?
A fly crawls across a hot plate (Exercise 257 ). Where it is, the plate warms by 4 °C per cm to the right and 3 °C per cm upward. The fly moves right at ¼ cm/s and up at ⅓ cm/s .
2 °C/s
4 · ¼ = 1 from moving right, 3 · ⅓ = 1 from moving up. Each direction the fly moves adds its own share of the warming.
time t (s) 3
Isotherms every 2 °C of a plate with T = 4 x + 3 y + 3 T=4x+3y+3 T = 4 x + 3 y + 3 x = 1 + t x=\sqrt{1+t} x = 1 + t y = 2 + t / 3 y=2+t/3 y = 2 + t /3
Two routes to one rate
Time moves the inputs; the inputs move the output. z z z depends on
t t t only through
x x x and
y y y — the
intermediate variables (
p362 ).
Each route is a product. (how fast z z z x x x x x x t t t , and the same through
y y y .
Add the routes. A tree diagram lists them: one branch per route (
p364 ).
Figure 4.34 (p365 ): multiply along a branch, add the branches.
One picture
Example 4.26a (p363 ): walk the unit circle x = sin t x=\sin t x = sin t y = cos t y=\cos t y = cos t z = 4 x 2 + 3 y 2 z=4x^2+3y^2 z = 4 x 2 + 3 y 2 z ( t ) = 3 + sin 2 t z(t)=3+\sin^2 t z ( t ) = 3 + sin 2 t
At t = π / 4 t=\pi/4 t = π /4 x-route gives + 4 +4 + 4 , the y-route gives − 3 -3 − 3 . You climb at 1 1 1
A rate along a path is the sum of the rates through each input.
Heights drawn at ⅓ scale. Drag to turn.
Level 2 · compute it Three chain rules and implicit slopes and three labs
One independent variable
Theorem 4.8 · p361
d z d t = ∂ z ∂ x d x d t + ∂ z ∂ y d y d t \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt} d t d z = ∂ x ∂ z d t d x + ∂ y ∂ z d t d y x = g ( t ) x=g(t) x = g ( t ) y = h ( t ) y=h(t) y = h ( t ) t t t ( x ( t ) , y ( t ) ) (x(t),y(t)) ( x ( t ) , y ( t ))
Example 4.26a · p363
d z d t = ( 8 x ) ( cos t ) + ( 6 y ) ( − sin t ) = 2 sin t cos t \frac{dz}{dt}=(8x)(\cos t)+(6y)(-\sin t)=2\sin t\cos t d t d z = ( 8 x ) ( cos t ) + ( 6 y ) ( − sin t ) = 2 sin t cos t Substitute x = sin t x=\sin t x = sin t y = cos t y=\cos t y = cos t 4 sin 2 t + 3 cos 2 t 4\sin^2t+3\cos^2t 4 sin 2 t + 3 cos 2 t
z z z t t t
Two independent variables
Theorem 4.9 · p365
∂ z ∂ u = ∂ z ∂ x ∂ x ∂ u + ∂ z ∂ y ∂ y ∂ u , ∂ z ∂ v = ∂ z ∂ x ∂ x ∂ v + ∂ z ∂ y ∂ y ∂ v \frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u},\qquad \frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial v} ∂ u ∂ z = ∂ x ∂ z ∂ u ∂ x + ∂ y ∂ z ∂ u ∂ y , ∂ v ∂ z = ∂ x ∂ z ∂ v ∂ x + ∂ y ∂ z ∂ v ∂ y x = g ( u , v ) x=g(u,v) x = g ( u , v ) y = h ( u , v ) y=h(u,v) y = h ( u , v ) ∂ z / ∂ u \partial z/\partial u ∂ z / ∂ u u u u
Example 4.27 · p366
∂ z ∂ u = 3 ( 6 x − 2 y ) + 4 ( − 2 x + 2 y ) = 38 u + 18 v \frac{\partial z}{\partial u}=3(6x-2y)+4(-2x+2y)=38u+18v ∂ u ∂ z = 3 ( 6 x − 2 y ) + 4 ( − 2 x + 2 y ) = 38 u + 18 v z = 3 x 2 − 2 x y + y 2 z=3x^2-2xy+y^2 z = 3 x 2 − 2 x y + y 2 x = 3 u + 2 v x=3u+2v x = 3 u + 2 v y = 4 u − v y=4u-v y = 4 u − v ∂ z / ∂ v = 18 u + 34 v \partial z/\partial v=18u+34v ∂ z / ∂ v = 18 u + 34 v
Now the left side is partial: z z z
Any number of variables
Theorem 4.10 · p367
∂ w ∂ t j = ∂ w ∂ x 1 ∂ x 1 ∂ t j + ∂ w ∂ x 2 ∂ x 2 ∂ t j + ⋯ + ∂ w ∂ x m ∂ x m ∂ t j \frac{\partial w}{\partial t_j}=\frac{\partial w}{\partial x_1}\frac{\partial x_1}{\partial t_j}+\frac{\partial w}{\partial x_2}\frac{\partial x_2}{\partial t_j}+\cdots+\frac{\partial w}{\partial x_m}\frac{\partial x_m}{\partial t_j} ∂ t j ∂ w = ∂ x 1 ∂ w ∂ t j ∂ x 1 + ∂ x 2 ∂ w ∂ t j ∂ x 2 + ⋯ + ∂ x m ∂ w ∂ t j ∂ x m One term per intermediate variable x i x_i x i t j t_j t j
Example 4.29 (p368 ): w = f ( x , y , z ) w=f(x,y,z) w = f ( x , y , z ) x , y , z x,y,z x , y , z t , u , v t,u,v t , u , v Count the leaves that end in your variable: that is the number of terms.
Implicit differentiation
Theorem 4.11 · p369
F ( x , y ) = 0 : d y d x = − F x F y ( F y ≠ 0 ) F(x,y)=0:\qquad \frac{dy}{dx}=-\frac{F_x}{F_y}\quad (F_y\neq0) F ( x , y ) = 0 : d x d y = − F y F x ( F y = 0 ) F ( x , y , z ) = 0 : ∂ z ∂ x = − F x F z , ∂ z ∂ y = − F y F z ( F z ≠ 0 ) F(x,y,z)=0:\qquad \frac{\partial z}{\partial x}=-\frac{F_x}{F_z},\quad \frac{\partial z}{\partial y}=-\frac{F_y}{F_z}\quad (F_z\neq0) F ( x , y , z ) = 0 : ∂ x ∂ z = − F z F x , ∂ y ∂ z = − F z F y ( F z = 0 ) Example 4.30a · p370
3 x 2 − 2 x y + y 2 + 4 x − 6 y − 11 = 0 : d y d x = 3 x − y + 2 x − y + 3 3x^2-2xy+y^2+4x-6y-11=0:\quad \frac{dy}{dx}=\frac{3x-y+2}{x-y+3} 3 x 2 − 2 x y + y 2 + 4 x − 6 y − 11 = 0 : d x d y = x − y + 3 3 x − y + 2 At ( 2 , 1 ) (2,1) ( 2 , 1 ) 7 4 \tfrac74 4 7 y = 7 4 x − 5 2 y=\tfrac74x-\tfrac52 y = 4 7 x − 2 5
Differentiate F F F x x x y y y
Examples 4.26–4.30
Example given result 4.26a · p363 z = 4 x 2 + 3 y 2 z=4x^2+3y^2 z = 4 x 2 + 3 y 2 x = sin t x=\sin t x = sin t y = cos t y=\cos t y = cos t d z d t = 2 sin t cos t \dfrac{dz}{dt}=2\sin t\cos t d t d z = 2 sin t cos t 4.26b · p364 z = x 2 − y 2 z=\sqrt{x^2-y^2} z = x 2 − y 2 x = e 2 t x=e^{2t} x = e 2 t y = e − t y=e^{-t} y = e − t d z d t = 2 e 6 t + 1 e t e 6 t − 1 \dfrac{dz}{dt}=\dfrac{2e^{6t}+1}{e^t\sqrt{e^{6t}-1}} d t d z = e t e 6 t − 1 2 e 6 t + 1 4.27 · p366 z = 3 x 2 − 2 x y + y 2 z=3x^2-2xy+y^2 z = 3 x 2 − 2 x y + y 2 x = 3 u + 2 v x=3u+2v x = 3 u + 2 v y = 4 u − v y=4u-v y = 4 u − v 38 u + 18 v 38u+18v 38 u + 18 v 18 u + 34 v \ 18u+34v 18 u + 34 v 4.28 · p367 w = 3 x 2 − 2 x y + 4 z 2 w=3x^2-2xy+4z^2 w = 3 x 2 − 2 x y + 4 z 2 x = e u sin v x=e^u\sin v x = e u sin v y = e u cos v y=e^u\cos v y = e u cos v z = e u z=e^u z = e u w u = 2 e 2 u ( 3 sin 2 v − 2 sin v cos v + 4 ) w_u=2e^{2u}(3\sin^2v-2\sin v\cos v+4) w u = 2 e 2 u ( 3 sin 2 v − 2 sin v cos v + 4 ) w v = 2 e 2 u ( sin 2 v + 3 sin v cos v − cos 2 v ) w_v=2e^{2u}(\sin^2v+3\sin v\cos v-\cos^2v) w v = 2 e 2 u ( sin 2 v + 3 sin v cos v − cos 2 v ) 4.29 · p368 w = f ( x , y , z ) w=f(x,y,z) w = f ( x , y , z ) x , y , z x,y,z x , y , z t , u , v t,u,v t , u , v 9 leaves; three sums of three products 4.30a · p370 3 x 2 − 2 x y + y 2 + 4 x − 6 y − 11 = 0 3x^2-2xy+y^2+4x-6y-11=0 3 x 2 − 2 x y + y 2 + 4 x − 6 y − 11 = 0 ( 2 , 1 ) (2,1) ( 2 , 1 ) slope 7 4 \tfrac74 4 7 y = 7 4 x − 5 2 y=\tfrac74x-\tfrac52 y = 4 7 x − 2 5 4.30b · p371 x 2 e y − y z e x = 0 x^2e^y-yze^x=0 x 2 e y − y z e x = 0 ∂ z ∂ x = 2 x e y − y z e x y e x \dfrac{\partial z}{\partial x}=\dfrac{2xe^y-yze^x}{ye^x} ∂ x ∂ z = y e x 2 x e y − y z e x ∂ z ∂ y = x 2 e y − z e x y e x \ \dfrac{\partial z}{\partial y}=\dfrac{x^2e^y-ze^x}{ye^x} ∂ y ∂ z = y e x x 2 e y − z e x
* The book's final line for ∂ w / ∂ v \partial w/\partial v ∂ w / ∂ v
Lab 1 · a point on a path
example
Ex 4.26a · bowl
Checkpoint 4.23
Exercise 257 · fly
time t 0.79
play t
Try: play the bowl. When the blue and amber terms cancel, the path runs along a contour and z(t) is flat.
Lab 2 · the tree diagram
case
x (t ), y (t )x (u , v ), y (u , v )Checkpoint 4.26
Example 4.29
Try: Example 4.29, then each of t, u, v. Every intermediate variable that depends on your variable contributes exactly one term.
Lab 3 · slopes from a level curve
curve
Figure 4.37
Example 4.30a
Checkpoint 4.27
Drag the point; it stays on the curve.
Try: drag to the far left or right of an ellipse. There F y = 0 F_y=0 F y = 0
Common mistakes
Mistake Result Fix Only one route 8 x cos t 8x\cos t 8 x cos t one term per branch: add ( 6 y ) ( − sin t ) (6y)(-\sin t) ( 6 y ) ( − sin t ) Answer left in x x x y y y 8 x cos t − 6 y sin t 8x\cos t-6y\sin t 8 x cos t − 6 y sin t t t t substitute x ( t ) x(t) x ( t ) y ( t ) y(t) y ( t ) d z / d u dz/du d z / d u hides that v v v ∂ z / ∂ u \partial z/\partial u ∂ z / ∂ u Partials evaluated at t t t f x ( t , t ) f_x(t,t) f x ( t , t ) f x ( x ( t ) , y ( t ) ) f_x(x(t),y(t)) f x ( x ( t ) , y ( t )) partials at the point on the path Dropping the minus in − F x / F y -F_x/F_y − F x / F y slope − 7 4 -\tfrac74 − 4 7 ( 2 , 1 ) (2,1) ( 2 , 1 ) F F F F y / F x F_y/F_x F y / F x slope 4 7 \tfrac47 7 4 7 4 \tfrac74 4 7 the variable you differentiate by is on top Misreading the tree a v v v ∂ z / ∂ u \partial z/\partial u ∂ z / ∂ u keep only leaves labelled with your variable
Practice and answers
Exercises from p371 ; the book's key: p862 –863 , p868 –869 .
Checkpoints 4.23–4.27
# task answer 4.23 z = x 2 − 3 x y + 2 y 2 z=x^2-3xy+2y^2 z = x 2 − 3 x y + 2 y 2 x = 3 sin 2 t x=3\sin2t x = 3 sin 2 t y = 4 cos 2 t y=4\cos2t y = 4 cos 2 t d z d t = − 46 sin 4 t − 72 cos 4 t \dfrac{dz}{dt}=-46\sin4t-72\cos4t d t d z = − 46 sin 4 t − 72 cos 4 t 4.24 z = 2 x − y x + 3 y z=\dfrac{2x-y}{x+3y} z = x + 3 y 2 x − y x = e 2 u cos 3 v x=e^{2u}\cos3v x = e 2 u cos 3 v y = e 2 u sin 3 v y=e^{2u}\sin3v y = e 2 u sin 3 v ∂ z ∂ u = 0 \dfrac{\partial z}{\partial u}=0 ∂ u ∂ z = 0 ∂ z ∂ v = − 21 ( 3 sin 3 v + cos 3 v ) 2 \ \dfrac{\partial z}{\partial v}=\dfrac{-21}{(3\sin3v+\cos3v)^2} ∂ v ∂ z = ( 3 sin 3 v + cos 3 v ) 2 − 21 4.25 w = x + 2 y − 4 z 2 x − y + 3 z w=\dfrac{x+2y-4z}{2x-y+3z} w = 2 x − y + 3 z x + 2 y − 4 z x = e 2 u cos 3 v x=e^{2u}\cos3v x = e 2 u cos 3 v y = e 2 u sin 3 v y=e^{2u}\sin3v y = e 2 u sin 3 v z = e 2 u z=e^{2u} z = e 2 u ∂ w ∂ u = 0 \dfrac{\partial w}{\partial u}=0 ∂ u ∂ w = 0 ∂ w ∂ v = 15 − 33 sin 3 v + 6 cos 3 v ( 3 + 2 cos 3 v − sin 3 v ) 2 \ \dfrac{\partial w}{\partial v}=\dfrac{15-33\sin3v+6\cos3v}{(3+2\cos3v-\sin3v)^2} ∂ v ∂ w = ( 3 + 2 cos 3 v − sin 3 v ) 2 15 − 33 sin 3 v + 6 cos 3 v 4.26 tree for w = f ( x , y ) w=f(x,y) w = f ( x , y ) x , y x,y x , y t , u , v t,u,v t , u , v 6 leaves; e.g. ∂ w ∂ t = ∂ w ∂ x ∂ x ∂ t + ∂ w ∂ y ∂ y ∂ t \dfrac{\partial w}{\partial t}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial t}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial t} ∂ t ∂ w = ∂ x ∂ w ∂ t ∂ x + ∂ y ∂ w ∂ t ∂ y 4.27 x 2 + x y − y 2 + 7 x − 3 y − 26 = 0 x^2+xy-y^2+7x-3y-26=0 x 2 + x y − y 2 + 7 x − 3 y − 26 = 0 ( 3 , − 2 ) (3,-2) ( 3 , − 2 ) d y d x = − 11 4 \dfrac{dy}{dx}=-\dfrac{11}{4} d x d y = − 4 11 y = − 11 4 x + 25 4 \ y=-\dfrac{11}{4}x+\dfrac{25}{4} y = − 4 11 x + 4 25
Exercises
# task answer key step 215 w = x y cos z w=xy\cos z w = x y cos z x = t x=t x = t y = t 2 y=t^2 y = t 2 z = arcsin t z=\arcsin t z = arcsin t 3 t 2 − 4 t 4 1 − t 2 \dfrac{3t^2-4t^4}{\sqrt{1-t^2}} 1 − t 2 3 t 2 − 4 t 4 cos ( arcsin t ) = 1 − t 2 \cos(\arcsin t)=\sqrt{1-t^2} cos ( arcsin t ) = 1 − t 2 217 w = 5 x 2 + 2 y 2 w=5x^2+2y^2 w = 5 x 2 + 2 y 2 x = − 3 s + t x=-3s+t x = − 3 s + t y = s − 4 t y=s-4t y = s − 4 t 94 s − 46 t 94s-46t 94 s − 46 t − 46 s + 74 t \ -46s+74t − 46 s + 74 t two branches each 219 f = x y f=xy f = x y x , y x,y x , y ∂ f / ∂ r \partial f/\partial r ∂ f / ∂ r r sin 2 θ r\sin2\theta r sin 2 θ 2 r sin θ cos θ 2r\sin\theta\cos\theta 2 r sin θ cos θ 221 f = x 2 + y 2 f=x^2+y^2 f = x 2 + y 2 x = t x=t x = t y = t 2 y=t^2 y = t 2 2 t + 4 t 3 2t+4t^3 2 t + 4 t 3 both methods agree 223 f = x y f=xy f = x y x = 1 − t x=1-\sqrt t x = 1 − t y = 1 + t y=1+\sqrt t y = 1 + t − 1 -1 − 1 f = 1 − t f=1-t f = 1 − t 227 w = x 2 + y 2 + z 2 w=x^2+y^2+z^2 w = x 2 + y 2 + z 2 x = cos t x=\cos t x = cos t y = sin t y=\sin t y = sin t z = e t z=e^t z = e t 2 e 2 t 2e^{2t} 2 e 2 t w = 1 + e 2 t w=1+e^{2t} w = 1 + e 2 t 229 u = e x sin y u=e^x\sin y u = e x sin y x = − ln 2 t x=-\ln2t x = − ln 2 t y = π t y=\pi t y = π t x = ln 2 x=\ln2 x = ln 2 2 ( π − 4 ) \sqrt2(\pi-4) 2 ( π − 4 ) t = 1 4 t=\tfrac14 t = 4 1 231 x 3 + y 2 x − 3 = 0 x^3+y^2x-3=0 x 3 + y 2 x − 3 = 0 − 3 x 2 + y 2 2 x y -\dfrac{3x^2+y^2}{2xy} − 2 x y 3 x 2 + y 2 − F x / F y -F_x/F_y − F x / F y 237 e x y + y e y = 1 e^{xy}+ye^y=1 e x y + y e y = 1 − y e x y x e x y + e y ( 1 + y ) -\dfrac{ye^{xy}}{xe^{xy}+e^y(1+y)} − x e x y + e y ( 1 + y ) y e x y product rule in F y F_y F y 243 z = x / y z=x/y z = x / y x = 2 cos u x=2\cos u x = 2 cos u y = 3 sin v y=3\sin v y = 3 sin v − 2 sin u 3 sin v \dfrac{-2\sin u}{3\sin v} 3 sin v − 2 sin u − 2 cos u cos v 3 sin 2 v \ \dfrac{-2\cos u\cos v}{3\sin^2v} 3 sin 2 v − 2 cos u cos v one branch each 245 z = x y e x / y z=xye^{x/y} z = x y e x / y r = 2 r=2 r = 2 θ = π / 6 \theta=\pi/6 θ = π /6 3 e 3 \sqrt3\,e^{\sqrt3} 3 e 3 ( 2 − 4 3 ) e 3 \ (2-4\sqrt3)e^{\sqrt3} ( 2 − 4 3 ) e 3 ( x , y ) = ( 3 , 1 ) (x,y)=(\sqrt3,1) ( x , y ) = ( 3 , 1 ) 251 cylinder V = π x 2 y V=\pi x^2y V = π x 2 y x = t / 2 x=t/2 x = t /2 y = t / 3 y=t/3 y = t /3 x = 2 x=2 x = 2 4 π 4\pi 4 π t = 4 t=4 t = 4 253 cone: r ′ = 3 r'=3 r ′ = 3 h ′ = − 2 h'=-2 h ′ = − 2 r = 13 r=13 r = 13 h = 18 h=18 h = 18 p373 ) 1066 π 3 ≈ 1116.31 \tfrac{1066\pi}{3}\approx1116.31 3 1066 π ≈ 1116.31 468 π − 338 π 3 468\pi-\tfrac{338\pi}{3} 468 π − 3 338 π 255 box surface area, each side + 0.5 +0.5 + 0.5 12 in²/min 2 ( x + y + z ) 2(x+y+z) 2 ( x + y + z ) 257 the fly 2 °C/s 4 ⋅ 1 4 + 3 ⋅ 1 3 4\cdot\tfrac14+3\cdot\tfrac13 4 ⋅ 4 1 + 3 ⋅ 3 1
Level 3 · why it works Where the rules come from short arguments
Why the routes add
The book's proof uses differentiability (p361 ). Near ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 )
z − z 0 = f x Δ x + f y Δ y + E , E Δ x 2 + Δ y 2 → 0 z-z_0=f_x\,\Delta x+f_y\,\Delta y+E,\qquad \frac{E}{\sqrt{\Delta x^2+\Delta y^2}}\to0 z − z 0 = f x Δ x + f y Δ y + E , Δ x 2 + Δ y 2 E → 0 z − z 0 Δ t = f x Δ x Δ t + f y Δ y Δ t + E Δ x 2 + Δ y 2 ⋅ Δ x 2 + Δ y 2 Δ t \frac{z-z_0}{\Delta t}=f_x\frac{\Delta x}{\Delta t}+f_y\frac{\Delta y}{\Delta t}+\frac{E}{\sqrt{\Delta x^2+\Delta y^2}}\cdot\frac{\sqrt{\Delta x^2+\Delta y^2}}{\Delta t} Δ t z − z 0 = f x Δ t Δ x + f y Δ t Δ y + Δ x 2 + Δ y 2 E ⋅ Δ t Δ x 2 + Δ y 2 As Δ t → 0 \Delta t\to0 Δ t → 0 0 × x ′ 2 + y ′ 2 = 0 0\times\sqrt{x'^2+y'^2}=0 0 × x ′2 + y ′2 = 0 Close up, the surface is its tangent plane, and on a plane the routes add exactly.
The chain rule as a dot product
d z d t = ⟨ f x , f y ⟩ ⋅ ⟨ x ′ ( t ) , y ′ ( t ) ⟩ \frac{dz}{dt}=\langle f_x,\ f_y\rangle\cdot\langle x'(t),\ y'(t)\rangle d t d z = ⟨ f x , f y ⟩ ⋅ ⟨ x ′ ( t ) , y ′ ( t )⟩ Example 4.26a at t = π / 4 t=\pi/4 t = π /4 ⟨ 5.66 , 4.24 ⟩ ⋅ ⟨ 0.71 , − 0.71 ⟩ = 1 \langle 5.66,\ 4.24\rangle\cdot\langle 0.71,\ -0.71\rangle=1 ⟨ 5.66 , 4.24 ⟩ ⋅ ⟨ 0.71 , − 0.71 ⟩ = 1 z z z Tick "show ∇f" in Lab 1 and watch the angle.
Where −Fx /Fy comes from
If F ( x , y ( x ) ) = 0 F(x,y(x))=0 F ( x , y ( x )) = 0 x x x t = x t=x t = x p370 )
F x ⋅ 1 + F y ⋅ d y d x = 0 ⟹ d y d x = − F x F y . F_x\cdot1+F_y\cdot\frac{dy}{dx}=0\quad\Longrightarrow\quad \frac{dy}{dx}=-\frac{F_x}{F_y}. F x ⋅ 1 + F y ⋅ d x d y = 0 ⟹ d x d y = − F y F x . With z ( x , y ) z(x,y) z ( x , y ) F ( x , y , z ) = 0 F(x,y,z)=0 F ( x , y , z ) = 0 y y y F x + F z ∂ z / ∂ x = 0 F_x+F_z\,\partial z/\partial x=0 F x + F z ∂ z / ∂ x = 0 The route through x x x y y y F F F
Why some routes cancel
In Checkpoint 4.24, u u u x x x y y y e 2 u e^{2u} e 2 u z = 2 x − y x + 3 y z=\dfrac{2x-y}{x+3y} z = x + 3 y 2 x − y x x x y y y ∂ z / ∂ u = 0 \partial z/\partial u=0 ∂ z / ∂ u = 0
Such an f f f homogeneous (p372 ): f ( k x , k y ) = k n f ( x , y ) f(kx,ky)=k^nf(x,y) f ( k x , k y ) = k n f ( x , y ) k k k k = 1 k=1 k = 1
x f x + y f y = n f x\,f_x+y\,f_y=n\,f x f x + y f y = n f For n = 0 n=0 n = 0 A zero in a chain-rule answer is often a symmetry, not luck.
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Calculus Volume 3 §4.5 Chapter 4 · Differentiation of Functions of Several Variables
The Chain Rule
When the inputs move, the output changes through every input at once.
One product per route from the output to the variable, then add: d z d t = ∂ z ∂ x d x d t + ∂ z ∂ y d y d t \dfrac{dz}{dt}=\dfrac{\partial z}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial z}{\partial y}\dfrac{dy}{dt} d t d z = ∂ x ∂ z d t d x + ∂ y ∂ z d t d y
4.5.1 State the chain rules for one or two independent variables. 4.5.2 Use tree diagrams as an aid to understanding the chain rule for several independent and intermediate variables. 4.5.3 Perform implicit differentiation of a function of two or more variables.
Ask: "One variable: d/dx f(g(x)) = ? What if f has two inputs that both move?"Prereq: d/dt sin(t²) · partials of x²y · implicit differentiation from Calc 1
Chain Rules for One or Two Independent Variables bridge Theorem 4.8
Moving along a path, the height changes through x x x y y y
x = g ( t ) , y = h ( t ) , z = f ( x , y ) x=g(t),\ y=h(t),\ z=f(x,y) x = g ( t ) , y = h ( t ) , z = f ( x , y ) d z d t = ∂ z ∂ x d x d t + ∂ z ∂ y d y d t \ \dfrac{dz}{dt}=\dfrac{\partial z}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial z}{\partial y}\dfrac{dy}{dt} d t d z = ∂ x ∂ z d t d x + ∂ y ∂ z d t d y
Across at d x d t \frac{dx}{dt} d t d x z z z ∂ z ∂ x \frac{\partial z}{\partial x} ∂ x ∂ z ∂ z ∂ x d x d t \ \frac{\partial z}{\partial x}\frac{dx}{dt} ∂ x ∂ z d t d x
Up at d y d t \frac{dy}{dt} d t d y ∂ z ∂ y d y d t \ \frac{\partial z}{\partial y}\frac{dy}{dt} ∂ y ∂ z d t d y
x , y x,y x , y intermediate variables — outputs of t t t f f f
Geometry split the velocity; each part crosses level curves at its own rate
Algebra · p361 f x x ′ ( t ) + f y y ′ ( t ) f_x\,x'(t)+f_y\,y'(t) f x x ′ ( t ) + f y y ′ ( t )
Ask: "Calc 1 multiplies. Two inputs: multiply what?"Partials at ( x , y ) (x,y) ( x , y ) t t t "Cancel" pattern: ∂ z ∂ x d x d t \frac{\partial z}{\partial x}\frac{dx}{dt} ∂ x ∂ z d t d x d z d t \frac{dz}{dt} d t d z
Chain Rules for One or Two Independent Variables algebra proof of Theorem 4.8
The routes add because, up close, a differentiable surface is its tangent plane
Differentiable: z − z 0 = f x ( x − x 0 ) + f y ( y − y 0 ) + E ( x , y ) z-z_0=f_x\,(x-x_0)+f_y\,(y-y_0)+E(x,y) z − z 0 = f x ( x − x 0 ) + f y ( y − y 0 ) + E ( x , y ) E ( x − x 0 ) 2 + ( y − y 0 ) 2 → 0 \dfrac{E}{\sqrt{(x-x_0)^2+(y-y_0)^2}}\to0 ( x − x 0 ) 2 + ( y − y 0 ) 2 E → 0 z ( t ) − z ( t 0 ) t − t 0 = f x x ( t ) − x ( t 0 ) t − t 0 + f y y ( t ) − y ( t 0 ) t − t 0 + E ( x ( t ) , y ( t ) ) t − t 0 \frac{z(t)-z(t_0)}{t-t_0}=f_x\frac{x(t)-x(t_0)}{t-t_0}+f_y\frac{y(t)-y(t_0)}{t-t_0}+\frac{E(x(t),y(t))}{t-t_0} t − t 0 z ( t ) − z ( t 0 ) = f x t − t 0 x ( t ) − x ( t 0 ) + f y t − t 0 y ( t ) − y ( t 0 ) + t − t 0 E ( x ( t ) , y ( t )) E t − t 0 = E ( x − x 0 ) 2 + ( y − y 0 ) 2 ⏟ → 0 ⋅ ( x − x 0 ) 2 + ( y − y 0 ) 2 t − t 0 ⏟ → ± x ′ ( t 0 ) 2 + y ′ ( t 0 ) 2 \frac{E}{t-t_0}=\underbrace{\frac{E}{\sqrt{(x-x_0)^2+(y-y_0)^2}}}_{\to\,0}\cdot\underbrace{\frac{\sqrt{(x-x_0)^2+(y-y_0)^2}}{t-t_0}}_{\to\,\pm\sqrt{x'(t_0)^2+y'(t_0)^2}} t − t 0 E = → 0 ( x − x 0 ) 2 + ( y − y 0 ) 2 E ⋅ → ± x ′ ( t 0 ) 2 + y ′ ( t 0 ) 2 t − t 0 ( x − x 0 ) 2 + ( y − y 0 ) 2 Let t → t 0 t\to t_0 t → t 0 d z d t = f x x ′ + f y y ′ \dfrac{dz}{dt}=f_x\,x'+f_y\,y' d t d z = f x x ′ + f y y ′
Only analysis step: E shrinks faster than the distance travelled.
Book p361–362: Equation 4.30, divide by t − t 0 t-t_0 t − t 0 Ask: "Why is the second limit finite?" → x, y differentiable in t
Chain Rules for One or Two Independent Variables geometry Example 4.26a
Walking the unit circle under z = 4 x 2 + 3 y 2 z=4x^2+3y^2 z = 4 x 2 + 3 y 2 2 sin t cos t 2\sin t\cos t 2 sin t cos t
x = sin t , y = cos t x=\sin t,\ y=\cos t x = sin t , y = cos t d z d t = 8 x cos t − 6 y sin t = 2 sin t cos t \ \dfrac{dz}{dt}=8x\cos t-6y\sin t=2\sin t\cos t d t d z = 8 x cos t − 6 y sin t = 2 sin t cos t
Lifted onto the bowl: height z ( t ) = 3 + sin 2 t z(t)=3+\sin^2 t z ( t ) = 3 + sin 2 t
At t = π 4 t=\tfrac\pi4 t = 4 π x x x 8 x cos t = 4 8x\cos t=4 8 x cos t = 4 y y y − 6 y sin t = − 3 -6y\sin t=-3 − 6 y sin t = − 3
Climbing at 4 − 3 = 1 4-3=1 4 − 3 = 1
Geometry slope of the lifted circle
Algebra · p363 8 x cos t − 6 y sin t 8x\cos t-6y\sin t 8 x cos t − 6 y sin t
Heights drawn at ⅓ scale; drag the bowl after the last click Ask: "Where on the lap is the height not changing?" → t = 0, π/2, π, 3π/2
Chain Rules for One or Two Independent Variables algebra Example 4.26a
Four derivatives, two products, then write the answer in t t t
Find ∂ z ∂ x , ∂ z ∂ y , d x d t , d y d t \frac{\partial z}{\partial x},\ \frac{\partial z}{\partial y},\ \frac{dx}{dt},\ \frac{dy}{dt} ∂ x ∂ z , ∂ y ∂ z , d t d x , d t d y x ( t ) , y ( t ) x(t),\ y(t) x ( t ) , y ( t )
∂ z ∂ x = 8 x , ∂ z ∂ y = 6 y , d x d t = cos t , d y d t = − sin t \frac{\partial z}{\partial x}=8x,\ \ \frac{\partial z}{\partial y}=6y,\ \ \frac{dx}{dt}=\cos t,\ \ \frac{dy}{dt}=-\sin t ∂ x ∂ z = 8 x , ∂ y ∂ z = 6 y , d t d x = cos t , d t d y = − sin t
d z d t = ( 8 x ) ( cos t ) + ( 6 y ) ( − sin t ) \dfrac{dz}{dt}=(8x)(\cos t)+(6y)(-\sin t) d t d z = ( 8 x ) ( cos t ) + ( 6 y ) ( − sin t )
= 8 sin t cos t − 6 cos t sin t = =8\sin t\cos t-6\cos t\sin t= = 8 sin t cos t − 6 cos t sin t = 2 sin t cos t 2\sin t\cos t 2 sin t cos t
Check: z = 4 sin 2 t + 3 cos 2 t z=4\sin^2t+3\cos^2t z = 4 sin 2 t + 3 cos 2 t d z d t \frac{dz}{dt} d t d z
Whiteboards: the four quantities before click 1 Slip: stopping at 8x cos t − 6y sin t Substituting first works here; often it doesn't
Chain Rules for One or Two Independent Variables algebra Example 4.26b · Checkpoint 4.23
With x = e 2 t , y = e − t x=e^{2t},\ y=e^{-t} x = e 2 t , y = e − t d d t x 2 − y 2 = 2 e 6 t + 1 e t e 6 t − 1 \ \frac{d}{dt}\sqrt{x^2-y^2}=\frac{2e^{6t}+1}{e^t\sqrt{e^{6t}-1}} d t d x 2 − y 2 = e t e 6 t − 1 2 e 6 t + 1
Same recipe; simplify only after substituting x ( t ) , y ( t ) x(t),\ y(t) x ( t ) , y ( t )
z = x 2 − y 2 , x = e 2 t , y = e − t z=\sqrt{x^2-y^2},\ \ x=e^{2t},\ \ y=e^{-t} z = x 2 − y 2 , x = e 2 t , y = e − t
d z d t = x x 2 − y 2 ( 2 e 2 t ) + − y x 2 − y 2 ( − e − t ) \dfrac{dz}{dt}=\dfrac{x}{\sqrt{x^2-y^2}}\,(2e^{2t})+\dfrac{-y}{\sqrt{x^2-y^2}}\,(-e^{-t}) d t d z = x 2 − y 2 x ( 2 e 2 t ) + x 2 − y 2 − y ( − e − t )
= 2 x e 2 t + y e − t x 2 − y 2 =\dfrac{2xe^{2t}+ye^{-t}}{\sqrt{x^2-y^2}} = x 2 − y 2 2 x e 2 t + y e − t
= 2 e 6 t + 1 e t e 6 t − 1 =\frac{2e^{6t}+1}{e^t\sqrt{e^{6t}-1}} = e t e 6 t − 1 2 e 6 t + 1
Checkpoint 4.23 z = x 2 − 3 x y + 2 y 2 z=x^2-3xy+2y^2 z = x 2 − 3 x y + 2 y 2 x = 3 sin 2 t x=3\sin2t x = 3 sin 2 t y = 4 cos 2 t y=4\cos2t y = 4 cos 2 t d z d t \frac{dz}{dt} d t d z t t t
− 46 sin 4 t − 72 cos 4 t -46\sin4t-72\cos4t − 46 sin 4 t − 72 cos 4 t
Book multiplies top by e 2 t e^{2t} e 2 t e 4 t \sqrt{e^{4t}} e 4 t CP 4.23: z x = 2 x − 3 y , z y = − 3 x + 4 y z_x=2x-3y,\ z_y=-3x+4y z x = 2 x − 3 y , z y = − 3 x + 4 y
Chain Rules for One or Two Independent Variables bridge Figure 4.34
A tree diagram: multiply along each branch, add the branches
Edge = derivative of the upper variable by the lower one. One branch to t t t
Branch 1: z → x → t z\to x\to t z → x → t ∂ z ∂ x d x d t \ \frac{\partial z}{\partial x}\frac{dx}{dt} ∂ x ∂ z d t d x
Branch 2: z → y → t z\to y\to t z → y → t ∂ z ∂ y d y d t \ \frac{\partial z}{\partial y}\frac{dy}{dt} ∂ y ∂ z d t d y
Geometry two paths from
z z z to
t t t Algebra · p365 two terms in (4.29)
Ask: "How many terms if z = f(x, y, w), all of t?" → 3LO 4.5.2
Chain Rules for One or Two Independent Variables bridge Theorem 4.9
Two independent variables: follow only the branches that end in your variable
x = g ( u , v ) , y = h ( u , v ) x=g(u,v),\ y=h(u,v) x = g ( u , v ) , y = h ( u , v ) ∂ z ∂ u = ∂ z ∂ x ∂ x ∂ u + ∂ z ∂ y ∂ y ∂ u \ \dfrac{\partial z}{\partial u}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u} ∂ u ∂ z = ∂ x ∂ z ∂ u ∂ x + ∂ y ∂ z ∂ u ∂ y v v v
Branches ending in u u u
Branches ending in v v v
∂ z / ∂ u \partial z/\partial u ∂ z / ∂ u d z / d u dz/du d z / d u z z z v v v
Geometry 4 leaves, 2 end in
u u u Algebra · p365 2 terms in
∂ z ∂ u \frac{\partial z}{\partial u} ∂ u ∂ z
Ask: "Why a partial on the left now?" (p366)Slip: writing dz/du
Chain Rules for One or Two Independent Variables algebra Example 4.27
Six partials, two sums: ∂ z ∂ u = 38 u + 18 v \frac{\partial z}{\partial u}=38u+18v ∂ u ∂ z = 38 u + 18 v ∂ z ∂ v = 18 u + 34 v \ \frac{\partial z}{\partial v}=18u+34v ∂ v ∂ z = 18 u + 34 v
List all six partials first; substitute x ( u , v ) , y ( u , v ) x(u,v),\ y(u,v) x ( u , v ) , y ( u , v )
z = 3 x 2 − 2 x y + y 2 , x = 3 u + 2 v , y = 4 u − v z=3x^2-2xy+y^2,\ \ x=3u+2v,\ \ y=4u-v z = 3 x 2 − 2 x y + y 2 , x = 3 u + 2 v , y = 4 u − v
z x = 6 x − 2 y , z y = − 2 x + 2 y x u = 3 , x v = 2 y u = 4 , y v = − 1 \begin{aligned}z_x&=6x-2y, & z_y&=-2x+2y\\ x_u&=3, & x_v&=2\\ y_u&=4, & y_v&=-1\end{aligned} z x x u y u = 6 x − 2 y , = 3 , = 4 , z y x v y v = − 2 x + 2 y = 2 = − 1
∂ z ∂ u = 3 ( 6 x − 2 y ) + 4 ( − 2 x + 2 y ) = 10 x + 2 y = 38 u + 18 v \begin{aligned}\frac{\partial z}{\partial u}&=3(6x-2y)+4(-2x+2y)\\&=10x+2y=38u+18v\end{aligned} ∂ u ∂ z = 3 ( 6 x − 2 y ) + 4 ( − 2 x + 2 y ) = 10 x + 2 y = 38 u + 18 v ∂ z ∂ v = 2 ( 6 x − 2 y ) − ( − 2 x + 2 y ) = 14 x − 6 y = 18 u + 34 v \begin{aligned}\frac{\partial z}{\partial v}&=2(6x-2y)-(-2x+2y)\\&=14x-6y=18u+34v\end{aligned} ∂ v ∂ z = 2 ( 6 x − 2 y ) − ( − 2 x + 2 y ) = 14 x − 6 y = 18 u + 34 v
Fast group: substitute first and compare Slip: sign of y v = − 1 y_v=-1 y v = − 1
Chain Rules for One or Two Independent Variables bridge Checkpoint 4.24
∂ z / ∂ u = 0 \partial z/\partial u=0 ∂ z / ∂ u = 0 u u u z z z A path along a level set of z z z
z = 2 x − y x + 3 y , x = e 2 u cos 3 v , y = e 2 u sin 3 v z=\dfrac{2x-y}{x+3y},\ \ x=e^{2u}\cos3v,\ \ y=e^{2u}\sin3v z = x + 3 y 2 x − y , x = e 2 u cos 3 v , y = e 2 u sin 3 v
u u u e 2 u e^{2u} e 2 u
… and z z z ∂ z ∂ u = 0 \frac{\partial z}{\partial u}=0 ∂ u ∂ z = 0
v v v ∂ z ∂ v = − 21 ( 3 sin 3 v + cos 3 v ) 2 \dfrac{\partial z}{\partial v}=\dfrac{-21}{(3\sin3v+\cos3v)^2} ∂ v ∂ z = ( 3 sin 3 v + cos 3 v ) 2 − 21
Geometry path along a level set
Algebra · p366 chain-rule terms cancel to 0
Ask: "Before computing: which derivative must be 0?"Level sets: 2 x − y = c ( x + 3 y ) 2x-y=c(x+3y) 2 x − y = c ( x + 3 y ) Exercises 248–250: x f x + y f y = n f xf_x+yf_y=nf x f x + y f y = n f n = 0 n=0 n = 0
The Generalized Chain Rule bridge Theorem 4.10
Any number of variables: one term for every intermediate variable
w = f ( x 1 , … , x m ) w=f(x_1,\dots,x_m) w = f ( x 1 , … , x m ) x i ( t 1 , … , t n ) x_i(t_1,\dots,t_n) x i ( t 1 , … , t n ) ∂ w ∂ t j = ∂ w ∂ x 1 ∂ x 1 ∂ t j + ⋯ + ∂ w ∂ x m ∂ x m ∂ t j \ \dfrac{\partial w}{\partial t_j}=\dfrac{\partial w}{\partial x_1}\dfrac{\partial x_1}{\partial t_j}+\dots+\dfrac{\partial w}{\partial x_m}\dfrac{\partial x_m}{\partial t_j} ∂ t j ∂ w = ∂ x 1 ∂ w ∂ t j ∂ x 1 + ⋯ + ∂ x m ∂ w ∂ t j ∂ x m
m = 3 m=3 m = 3 x , y , z x,y,z x , y , z n = 2 n=2 n = 2 u , v u,v u , v
∂ w ∂ u \frac{\partial w}{\partial u} ∂ u ∂ w u u u
∂ w ∂ v \frac{\partial w}{\partial v} ∂ v ∂ w v v v
Geometry m m m branches end in
t j t_j t j Algebra · p367 m m m products in (4.33)
Ask: "Terms in ∂w/∂u? Partials to compute in total?" → 3; 9This tree = Example 4.28's structure
The Generalized Chain Rule algebra Example 4.28 · Checkpoint 4.25
Nine partials give ∂ w ∂ u = 2 e 2 u ( 3 sin 2 v − 2 sin v cos v + 4 ) \frac{\partial w}{\partial u}=2e^{2u}(3\sin^2v-2\sin v\cos v+4) ∂ u ∂ w = 2 e 2 u ( 3 sin 2 v − 2 sin v cos v + 4 )
∂ w ∂ u = w x x u + w y y u + w z z u \dfrac{\partial w}{\partial u}=w_xx_u+w_yy_u+w_zz_u ∂ u ∂ w = w x x u + w y y u + w z z u ∂ w ∂ v = w x x v + w y y v + w z z v \ \dfrac{\partial w}{\partial v}=w_xx_v+w_yy_v+w_zz_v ∂ v ∂ w = w x x v + w y y v + w z z v
w = 3 x 2 − 2 x y + 4 z 2 , x = e u sin v , y = e u cos v , z = e u w=3x^2-2xy+4z^2,\ \ x=e^u\sin v,\ \ y=e^u\cos v,\ \ z=e^u w = 3 x 2 − 2 x y + 4 z 2 , x = e u sin v , y = e u cos v , z = e u
w x = 6 x − 2 y w y = − 2 x w z = 8 z x u = e u sin v y u = e u cos v z u = e u x v = e u cos v y v = − e u sin v z v = 0 \begin{aligned}w_x&=6x-2y & w_y&=-2x & w_z&=8z\\ x_u&=e^u\sin v & y_u&=e^u\cos v & z_u&=e^u\\ x_v&=e^u\cos v & y_v&=-e^u\sin v & z_v&=0\end{aligned} w x x u x v = 6 x − 2 y = e u sin v = e u cos v w y y u y v = − 2 x = e u cos v = − e u sin v w z z u z v = 8 z = e u = 0
∂ w ∂ u = 2 e 2 u ( 3 sin 2 v − 2 sin v cos v + 4 ) \frac{\partial w}{\partial u}=2e^{2u}\,(3\sin^2v-2\sin v\cos v+4) ∂ u ∂ w = 2 e 2 u ( 3 sin 2 v − 2 sin v cos v + 4 ) ∂ w ∂ v = 2 e 2 u ( sin 2 v + 3 sin v cos v − cos 2 v ) \frac{\partial w}{\partial v}=2e^{2u}\,(\sin^2v+3\sin v\cos v-\cos^2v) ∂ v ∂ w = 2 e 2 u ( sin 2 v + 3 sin v cos v − cos 2 v ) The book's last line for ∂ w / ∂ v \partial w/\partial v ∂ w / ∂ v
Checkpoint 4.25 w = x + 2 y − 4 z 2 x − y + 3 z w=\dfrac{x+2y-4z}{2x-y+3z} w = 2 x − y + 3 z x + 2 y − 4 z x = e 2 u cos 3 v , y = e 2 u sin 3 v , z = e 2 u x=e^{2u}\cos3v,\ y=e^{2u}\sin3v,\ z=e^{2u} x = e 2 u cos 3 v , y = e 2 u sin 3 v , z = e 2 u
∂ w ∂ u = 0 , ∂ w ∂ v = 15 − 33 sin 3 v + 6 cos 3 v ( 3 + 2 cos 3 v − sin 3 v ) 2 \dfrac{\partial w}{\partial u}=0,\quad \dfrac{\partial w}{\partial v}=\dfrac{15-33\sin3v+6\cos3v}{(3+2\cos3v-\sin3v)^2} ∂ u ∂ w = 0 , ∂ v ∂ w = ( 3 + 2 cos 3 v − sin 3 v ) 2 15 − 33 sin 3 v + 6 cos 3 v
Hand out as written work; don't do every line live z v = 0 z_v=0 z v = 0 CP 4.25: ∂w/∂u = 0, same reason as CP 4.24 (u only scales x, y, z)
The Generalized Chain Rule bridge Example 4.29 · Checkpoint 4.26
Three intermediate × three independent: nine leaves, three formulas of three terms
w = f ( x , y , z ) w=f(x,y,z) w = f ( x , y , z ) x , y , z x,y,z x , y , z ( t , u , v ) (t,u,v) ( t , u , v ) ∂ w ∂ t = ∂ w ∂ x ∂ x ∂ t + ∂ w ∂ y ∂ y ∂ t + ∂ w ∂ z ∂ z ∂ t \ \dfrac{\partial w}{\partial t}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial t}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial t}+\dfrac{\partial w}{\partial z}\dfrac{\partial z}{\partial t} ∂ t ∂ w = ∂ x ∂ w ∂ t ∂ x + ∂ y ∂ w ∂ t ∂ y + ∂ z ∂ w ∂ t ∂ z u , v u,v u , v
Branches ending in t t t
… in u u u
… in v v v
Checkpoint 4.26 w = f ( x , y ) w=f(x,y) w = f ( x , y ) x , y x,y x , y ( t , u , v ) (t,u,v) ( t , u , v )
6 leaves: ∂ w ∂ t = ∂ w ∂ x ∂ x ∂ t + ∂ w ∂ y ∂ y ∂ t \frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t} ∂ t ∂ w = ∂ x ∂ w ∂ t ∂ x + ∂ y ∂ w ∂ t ∂ y u , v u,v u , v
Ask: "Terms per formula = ?" → number of intermediate variablesCP 4.26: pairs draw the tree first
Implicit Differentiation bridge Theorem 4.11
On the curve f ( x , y ) = 0 f(x,y)=0 f ( x , y ) = 0 − f x / f y -f_x/f_y − f x / f y y y y
d y d x = − f x f y ( f y ≠ 0 ) \dfrac{dy}{dx}=-\dfrac{f_x}{f_y}\ (f_y\neq0) d x d y = − f y f x ( f y = 0 ) f ( x , y , z ) = 0 f(x,y,z)=0 f ( x , y , z ) = 0 ∂ z ∂ x = − f x f z , ∂ z ∂ y = − f y f z ( f z ≠ 0 ) \ \dfrac{\partial z}{\partial x}=-\dfrac{f_x}{f_z},\ \dfrac{\partial z}{\partial y}=-\dfrac{f_y}{f_z}\ (f_z\neq0) ∂ x ∂ z = − f z f x , ∂ y ∂ z = − f z f y ( f z = 0 )
x 2 + 3 y 2 + 4 y − 4 = 0 x^2+3y^2+4y-4=0 x 2 + 3 y 2 + 4 y − 4 = 0
Calculus 1: 2 x + 6 y y ′ + 4 y ′ = 0 ⇒ y ′ = − x 3 y + 2 2x+6y\,y'+4y'=0\ \Rightarrow\ y'=-\dfrac{x}{3y+2} 2 x + 6 y y ′ + 4 y ′ = 0 ⇒ y ′ = − 3 y + 2 x
Theorem 4.11: − f x f y = − 2 x 6 y + 4 = − x 3 y + 2 -\dfrac{f_x}{f_y}=-\dfrac{2x}{6y+4}=-\dfrac{x}{3y+2} − f y f x = − 6 y + 4 2 x = − 3 y + 2 x
At ( 2 , 0 ) (2,0) ( 2 , 0 ) − 1 -1 − 1
Geometry tangent to
f = 0 f=0 f = 0 , ⟂
⟨ f x , f y ⟩ \langle f_x,f_y\rangle ⟨ f x , f y ⟩ Algebra · p369 d y d x = − f x f y \frac{dy}{dx}=-\frac{f_x}{f_y} d x d y = − f y f x
Board: Calc 1 way first, then the one-liner Ask: "Where does the formula fail?" → f y = 0 f_y=0 f y = 0 LO 4.5.3
Implicit Differentiation algebra why Theorem 4.11 holds
Implicit differentiation is the chain rule applied to f ( x , y ( x ) ) = 0 f(x,y(x))=0 f ( x , y ( x )) = 0
d d x f ( x , y ( x ) ) = f x ⋅ 1 + f y d y d x = 0 ⇒ d y d x = − f x f y \dfrac{d}{dx}f(x,y(x))=f_x\cdot1+f_y\dfrac{dy}{dx}=0\ \Rightarrow\ \dfrac{dy}{dx}=-\dfrac{f_x}{f_y} d x d f ( x , y ( x )) = f x ⋅ 1 + f y d x d y = 0 ⇒ d x d y = − f y f x
Differentiate both sides of f ( x , y ) = 0 f(x,y)=0 f ( x , y ) = 0 x x x
Two branches end in x x x ∂ f ∂ x d x d x + ∂ f ∂ y d y d x = 0 \ \frac{\partial f}{\partial x}\frac{dx}{dx}+\frac{\partial f}{\partial y}\frac{dy}{dx}=0 ∂ x ∂ f d x d x + ∂ y ∂ f d x d y = 0
d x d x = 1 \frac{dx}{dx}=1 d x d x = 1 d y d x \frac{dy}{dx} d x d y
f ( x , y , z ( x , y ) ) = 0 f(x,y,z(x,y))=0 f ( x , y , z ( x , y )) = 0 f x + f z z x = 0 \ f_x+f_z\,z_x=0 f x + f z z x = 0
Book p370: Equation 4.34 from 4.31 Ask: "Why is the right side 0?" → f is constant (0) along the curve
Implicit Differentiation bridge Example 4.30a
At ( 2 , 1 ) (2,1) ( 2 , 1 ) 7 4 \tfrac74 4 7 y = 7 4 x − 5 2 y=\tfrac74x-\tfrac52 y = 4 7 x − 2 5
Tangent line: m = − f x f y m=-\dfrac{f_x}{f_y} m = − f y f x y − y 0 = m ( x − x 0 ) y-y_0=m(x-x_0) y − y 0 = m ( x − x 0 )
3 x 2 − 2 x y + y 2 + 4 x − 6 y − 11 = 0 3x^2-2xy+y^2+4x-6y-11=0 3 x 2 − 2 x y + y 2 + 4 x − 6 y − 11 = 0
f x = 6 x − 2 y + 4 = 14 f_x=6x-2y+4=14 f x = 6 x − 2 y + 4 = 14 f y = − 2 x + 2 y − 6 = − 8 f_y=-2x+2y-6=-8 f y = − 2 x + 2 y − 6 = − 8
d y d x = 3 x − y + 2 x − y + 3 \dfrac{dy}{dx}=\dfrac{3x-y+2}{x-y+3} d x d y = x − y + 3 3 x − y + 2 ( 2 , 1 ) (2,1) ( 2 , 1 ) 7 4 \tfrac74 4 7
y − 1 = 7 4 ( x − 2 ) ⇒ y = 7 4 x − 5 2 y-1=\tfrac74(x-2)\ \Rightarrow\ y=\tfrac74x-\tfrac52 y − 1 = 4 7 ( x − 2 ) ⇒ y = 4 7 x − 2 5
Drop the minus: slope − 7 4 -\tfrac74 − 4 7
Evaluate f_x, f_y at (2, 1) before simplifying Figure: ⟨14, −8⟩ is ⟂ to the tangent
Implicit Differentiation algebra Example 4.30b
For a surface f ( x , y , z ) = 0 f(x,y,z)=0 f ( x , y , z ) = 0 − f z -f_z − f z
∂ z ∂ x = − f x f z , ∂ z ∂ y = − f y f z ( f z ≠ 0 ) \dfrac{\partial z}{\partial x}=-\dfrac{f_x}{f_z},\quad \dfrac{\partial z}{\partial y}=-\dfrac{f_y}{f_z}\quad(f_z\neq0) ∂ x ∂ z = − f z f x , ∂ y ∂ z = − f z f y ( f z = 0 ) f ( x , y , z ) = x 2 e y − y z e x f(x,y,z)=x^2e^y-yze^x f ( x , y , z ) = x 2 e y − y z e x
f x = 2 x e y − y z e x , f y = x 2 e y − z e x , f z = − y e x f_x=2xe^y-yze^x,\quad f_y=x^2e^y-ze^x,\quad f_z=-ye^x f x = 2 x e y − y z e x , f y = x 2 e y − z e x , f z = − y e x
∂ z ∂ x = − 2 x e y − y z e x − y e x = 2 x e y − y z e x y e x \frac{\partial z}{\partial x}=-\frac{2xe^y-yze^x}{-ye^x}=\frac{2xe^y-yze^x}{ye^x} ∂ x ∂ z = − − y e x 2 x e y − y z e x = y e x 2 x e y − y z e x ∂ z ∂ y = − x 2 e y − z e x − y e x = x 2 e y − z e x y e x \frac{\partial z}{\partial y}=-\frac{x^2e^y-ze^x}{-ye^x}=\frac{x^2e^y-ze^x}{ye^x} ∂ y ∂ z = − − y e x x 2 e y − z e x = y e x x 2 e y − z e x
Ask: "Which two minus signs cancel?"Needs y ≠ 0 y\neq0 y = 0
Implicit Differentiation bridge Checkpoint 4.27
The tangent at ( 3 , − 2 ) (3,-2) ( 3 , − 2 ) y = − 11 4 x + 25 4 y=-\frac{11}{4}x+\frac{25}{4} y = − 4 11 x + 4 25
d y d x = − f x f y = 2 x + y + 7 2 y − x + 3 \dfrac{dy}{dx}=-\dfrac{f_x}{f_y}=\dfrac{2x+y+7}{2y-x+3} d x d y = − f y f x = 2 y − x + 3 2 x + y + 7
x 2 + x y − y 2 x^2+xy-y^2 x 2 + x y − y 2 + 7 x − 3 y − 26 = 0 \quad+7x-3y-26=0 + 7 x − 3 y − 26 = 0
f x = 2 x + y + 7 = 11 , f y = x − 2 y − 3 = 4 f_x=2x+y+7=11,\ \ f_y=x-2y-3=4 f x = 2 x + y + 7 = 11 , f y = x − 2 y − 3 = 4
slope − 11 4 -\tfrac{11}{4} − 4 11
y + 2 = − 11 4 ( x − 3 ) ⇒ y = − 11 4 x + 25 4 y+2=-\tfrac{11}{4}(x-3)\ \Rightarrow\ y=-\tfrac{11}{4}x+\tfrac{25}{4} y + 2 = − 4 11 ( x − 3 ) ⇒ y = − 4 11 x + 4 25
Pairs first; reveal per click Check (3, −2) is on the curve: 9 − 6 − 4 + 21 + 6 − 26 = 0
Section 4.5 Exercises bridge Exercise 257
A fly on a hot plate warms at 4 ⋅ 1 4 + 3 ⋅ 1 3 = 2 4\cdot\tfrac14+3\cdot\tfrac13=2 4 ⋅ 4 1 + 3 ⋅ 3 1 = 2
d T d t = T x d x d t + T y d y d t \dfrac{dT}{dt}=T_x\dfrac{dx}{dt}+T_y\dfrac{dy}{dt} d t d T = T x d t d x + T y d t d y
x = 1 + t , y = 2 + t 3 x=\sqrt{1+t},\ y=2+\tfrac t3 x = 1 + t , y = 2 + 3 t t = 3 t=3 t = 3 ( 2 , 3 ) (2,3) ( 2 , 3 ) T x = 4 , T y = 3 T_x=4,\ T_y=3 T x = 4 , T y = 3
Right at x ′ = 1 4 x'=\tfrac14 x ′ = 4 1 4 ⋅ 1 4 = 1 4\cdot\tfrac14=1 4 ⋅ 4 1 = 1
Up at y ′ = 1 3 y'=\tfrac13 y ′ = 3 1 3 ⋅ 1 3 = 1 3\cdot\tfrac13=1 3 ⋅ 3 1 = 1
d T d t = 1 + 1 = \dfrac{dT}{dt}=1+1= d t d T = 1 + 1 = 2 °C/s
Geometry velocity split right and up
Algebra · p373 T x x ′ + T y y ′ T_x\,x'+T_y\,y' T x x ′ + T y y ′
Ask for a guess first: many multiply, few addIsotherms drawn straight near (2, 3) (linear approximation)
Section 4.5 Exercises bridge Exercise 253
Wider beats shorter: the cone gains 1066 π 3 \tfrac{1066\pi}{3} 3 1066 π
V = 1 3 π r 2 h V=\tfrac13\pi r^2h V = 3 1 π r 2 h d V d t = 2 3 π r h d r d t + 1 3 π r 2 d h d t \ \dfrac{dV}{dt}=\tfrac23\pi rh\,\dfrac{dr}{dt}+\tfrac13\pi r^2\dfrac{dh}{dt} d t d V = 3 2 π r h d t d r + 3 1 π r 2 d t d h
r = 13 , h = 18 , r ′ = 3 , h ′ = − 2 r=13,\ h=18,\ r'=3,\ h'=-2 r = 13 , h = 18 , r ′ = 3 , h ′ = − 2
radius route: 2 3 π ( 13 ) ( 18 ) ⋅ 3 = 468 π \tfrac23\pi(13)(18)\cdot3=468\pi 3 2 π ( 13 ) ( 18 ) ⋅ 3 = 468 π
height route: 1 3 π ( 13 ) 2 ⋅ ( − 2 ) = − 338 π 3 \tfrac13\pi(13)^2\cdot(-2)=-\tfrac{338\pi}{3} 3 1 π ( 13 ) 2 ⋅ ( − 2 ) = − 3 338 π
d V d t = 1066 π 3 ≈ \dfrac{dV}{dt}=\dfrac{1066\pi}{3}\approx d t d V = 3 1066 π ≈ 1116.31 cm³/min
Same pattern as the fly: one product per changing input Units: cm² · cm/min = cm³/min
Looking ahead · §4.6 bridge
The chain rule is a dot product: ⟨ f x , f y ⟩ ⋅ ⟨ x ′ , y ′ ⟩ \langle f_x,f_y\rangle\cdot\langle x',y'\rangle ⟨ f x , f y ⟩ ⋅ ⟨ x ′ , y ′ ⟩
d z d t = ⟨ f x , f y ⟩ ⋅ ⟨ x ′ ( t ) , y ′ ( t ) ⟩ \dfrac{dz}{dt}=\langle f_x,\ f_y\rangle\cdot\langle x'(t),\ y'(t)\rangle d t d z = ⟨ f x , f y ⟩ ⋅ ⟨ x ′ ( t ) , y ′ ( t )⟩ ⟨ f x , f y ⟩ \langle f_x,f_y\rangle ⟨ f x , f y ⟩
Example 4.26a at t = π 4 t=\tfrac\pi4 t = 4 π ⟨ 5.66 , 4.24 ⟩ ⋅ ⟨ 0.71 , − 0.71 ⟩ = 1 \langle5.66,\,4.24\rangle\cdot\langle0.71,\,-0.71\rangle=1 ⟨ 5.66 , 4.24 ⟩ ⋅ ⟨ 0.71 , − 0.71 ⟩ = 1
Along a level curve d z / d t = 0 dz/dt=0 d z / d t = 0
That vector is the gradient — §4.6.
Same reason the tangent in implicit differentiation is ⟂ ⟨f_x, f_y⟩
§4.5 wrap-up
Count the routes, multiply along each, add them up
One term per intermediate variable · a partial on the left when other independent variables are held fixed · implicit: − f x / f y -f_x/f_y − f x / f y
Objective You can now
4.5.1 chain rules, one or two independent variablesd z / d t = z x x ′ + z y y ′ dz/dt=z_x\,x'+z_y\,y' d z / d t = z x x ′ + z y y ′ ∂ z / ∂ u = z x x u + z y y u \ \partial z/\partial u=z_x\,x_u+z_y\,y_u ∂ z / ∂ u = z x x u + z y y u 4.5.2 tree diagramsdraw the tree; one product per branch ending in the variable 4.5.3 implicit differentiationd y / d x = − f x / f y dy/dx=-f_x/f_y d y / d x = − f x / f y ∂ z / ∂ x = − f x / f z \ \partial z/\partial x=-f_x/f_z ∂ z / ∂ x = − f x / f z
Exit ticket: z = x y 2 , x = t 2 , y = sin t z=xy^2,\ x=t^2,\ y=\sin t z = x y 2 , x = t 2 , y = sin t d z / d t = 2 t sin 2 t + 2 t 2 sin t cos t dz/dt=2t\sin^2t+2t^2\sin t\cos t d z / d t = 2 t sin 2 t + 2 t 2 sin t cos t Homework: 215, 217, 219, 221, 223, 227, 229, 231, 237, 243, 245, 251, 253, 255, 257