Fwd: Ask My Instructor Question - Section 3.10, Question: 3.10.13

[William DeMeo]

---------- Forwarded message ----------
From: Ryan via Pearson Education <xl-no...@pearsoncmg.com>
Date: Mon, Mar 2, 2015 at 7:23 PM
Subject: Ask My Instructor Question - Section 3.10, Question: 3.10.13

Book title: Thomas' Calculus Early Transcendentals, 12e
Course name: MATH_-165_-14
Objective: Solve related rates problems.

Assign. 10 (Sec. 3.10, 3.11) (Question # 4)

Question Link (this link will remain valid for 6 months from 03/02/15):
http://www.mathxl.com/info/exercise.aspx?fromask=yes&dataid=0e216cdb-01ec-49d8-ac6b-71d5efc22312

After trying the problem once, it became easy to pick out what was
going to be the correct answer, however I don't understand why the
correct answer is so.

[William DeMeo]

The question gives you the equation for volume V of an object in terms of the objects dimensions, x and y, as follows:

V = 4 x^2 y

You are supposed to find dV/dt.  For part c, this is a simple application of the product and chain rules. (For the other parts, it's even easier.)  In part c, we assume that both x and y are functions of t, so we have a product of two functions.  Therefore, the product rule gives:

dV/dt = 4 d(x^2)/dt y + 4 x^2 dy/dt

Now use the chain rule to take the derivative of x^2, that is, d(x^2)/dt = 2x dx/dt.  Plug this into the previous expression to get the final answer.

-William

[William DeMeo]

---------- Forwarded message ----------
Date: Mon, Mar 2, 2015 at 8:17 PM
To: William DeMeo

I understand in part c: that  dV/dt = 8xy * dx/dt + 4x^2 * dy/dt

What threw me off was in parts a & b when they said that x & y were constants. Should I just be thinking that this implies dx/dt = 0 and dy/dt = 0?

[William DeMeo]

yes, when the problem states that x is
constant, that means x should be treated as a fixed number that
doesn't depend on t (so dx/dt = 0, as you noted).