1. ## Derivative application

A rectangular prism has its length increasing by 12 cm/min, its width increasing by 4 cm/min and its height increasing by 2 cm/min. How fast is it's volume changing when the dimensions are 200 cm in length, 50 cm in width and 30 cm in height?

The following is my working:

dL/dt = 12 cm/min
dW/dt = 4 cm/min
dH/dt = 2 cm/min

V = L x W x H
we get:
V=(200+12t)(50+4t)(30+3t)
$V = 144t^3 + 5460t^2 + 72000t + 300,000$
Derivative of V
$V' = 432t^2 + 10920t + 72000$

Now, I'm very confused. How could we find the rates of change,when 200 cm in length, 50 cm in width and 30 cm in height?

Thanks!

2. You need to take implicit differentiation of both sides of the volume equation:

V=L*W*H

dV/dt = (dL/dt)*W*H + L*(dW/dt)*H + .... (Use chain rule here)

Then all you need to do is sub in for expressions on the right side and you will get dV/dt which is what you're looking for.

Hope it helps

3. #1 - Get fluent at writing this sort of thing. It's just the product rule.

$\frac{d}{dt}[L(t)\cdot W(t) \cdot H(t)] = L'(t)\cdot [W(t) \cdot H(t)] + L(t)\cdot [W'(t)\cdot H(t) + W(t)\cdot H'(t)]$

#2 - Or, the Differential Version

$dA = [dL\cdot [W \cdot H] + L\cdot [dW\cdot H + W\cdot dH]]\cdot dt$

Now you fill in the known values:

$dA = 12\cdot [50 \cdot 30] + 200\cdot [4\cdot 30 + 50\cdot 2]$

Did I get them all in the right spots?

#3 - Really, get good at all three types of notation. Switch effortlessly between them. Let the notation help you.