Derivative application

May 2010
36
0
[FONT=&quot]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:

[/FONT] 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)
\(\displaystyle V = 144t^3 + 5460t^2 + 72000t + 300,000\)
Derivative of V
\(\displaystyle V' = 432t^2 + 10920t + 72000\)


Now, I'm very confused. How could we find the rates of change,when [FONT=&quot]200 cm in length, 50 cm in width and 30 cm in height?
[/FONT]
Thanks!


[FONT=&quot]
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May 2010
3
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
 
Aug 2007
3,171
860
USA
#1 - Get fluent at writing this sort of thing. It's just the product rule.

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

#2 - Or, the Differential Version

\(\displaystyle dA = [dL\cdot [W \cdot H] + L\cdot [dW\cdot H + W\cdot dH]]\cdot dt\)

Now you fill in the known values:

\(\displaystyle dA = 12\cdot [50 \cdot 30] + 200\cdot [4\cdot 30 + 50\cdot 2]\)

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.