Calc 1 review: Chain rule. Suppose z = f(y) and y = g(x). Then z = (f ∘ g)(x) \frac{dz}{dy} = f′(y) \frac{dy}{dx} = g′(x) \frac{dz}{dx} = (f ∘ g)′(x) Chain rule (f ∘ g)′(x) = f′(g(x))g′(x) \frac{dz}{dx} = (f ∘ g)′(x) = f′(g(x))g′(x) = \frac{dz}{dy} \frac{dy}{dx} Thus \frac{dz}{dx} = \frac{dz}{dy} \frac{dy}{dx}

Example: The area of a rectangle is given by the function A : R^{2} → R, A(l, w) = lw Suppose the length of a rectangle is changing with time according to the function l(t) = t^{2}. Suppose the width of a rectangle is changing with time according to the function w(t) = sin(t). We can combine these two functions for l and w into a function F : R → R^{2} where F (t) = Thus the area of a rectangle is changing with time according to the function:

At time t = 3, The length of the rectangle is The width of the rectangle is The area of the rectangle is The rate of change of the length of the rectangle at t = 3 is

The rate of change of the width of the rectangle at t = 3 is

The rate of change of the area of the rectangle at t = 3 is Example: The area of a rectangle is given by the function A : R^{2} → R, A(l, w) = lw

Suppose the length of a rectangle is changing with time according to some unknown function l : R → R

Suppose the width of a rectangle is changing with time according to some unknown function w : R → R

We can combine these two functions for l and w into a function F : R → R^{2} where F (t) =

Suppose at time t = 3, the length of the rectangle is 10m and the width is 8 m. Suppose also that at time t = 3, the length is decreasing at a rate of 4m/sec while the width is increasing at a rate of 5m/sec. Find the rate of change of the area of the rectangle at t = 3.