AP Calc AB questions

Test Preparation AP Mathematics & Computer Science
1

<p>Hi guys,</p>

<p>I have tried over and over but I still can't find out how to do these questions:
1. A cube is expanding so that its surface area is increasing at a constant rate of 4 square inches per second. How fast is the volume increasing at the instant when the surface area is 24 square inches?
answer is 2cubic in/sec</p>

<ol>
<li><p>What is the approximate slope of the tangent to the curve x^3+y^3=xy at x = 1? Calculator allowed
answer is -1.014 </p></li>
<li><p>Water is flowing into a spherical tank with a 6 foot radius at the constant rate of 30pi cubic feet per hour. When the water is h feet deep, the volume of water in the tank is given by the formula:V= ((pih^2)/3)(18-h)
Determine the rate (in ft per hr) that the depth of the water in the tank is increasing at the instant when the water is 2 feet deep.
answer is 1.5</p></li>
</ol>

<p>For questions 2 and 3 I'm allowed to use a calculator but I really don't know how I can use it for these questions.</p>

<p>Please help!</p>

<p>Thanks!!!</p>

<p>Sorry, I just figured out that I forgot to put the pi in question 3 when I was doing it, so i couldn't get the right answer. But I'm still having trouble with question 1 and 2 please help!</p>

2

<ol>
<li>A cube is expanding so that its surface area is increasing at a constant rate of 4 square inches per second. How fast is the volume increasing at the instant when the surface area is 24 square inches?
answer is 2cubic in/sec</li>
</ol>

<p>Chain rule:</p>

<p>dV/dt = dV/dA * dA/dt</p>

<p>dV/dt = ?
dA/dt = 4 in^2/second</p>

<p>to find dV/dA, we need volume in terms of surface area</p>

<p>V = a^3
A = 6a^2 —> a = sqrt(A/6)</p>

<p>so,
V = (A/6)^(3/2)</p>

<p>dV/dA = 1/6 * 3/2 * (A/6)^(1/2)</p>

<p>when surface area = 24 inches, dV/dA = 1/2</p>

<p>dV/dt= dV/dA * dA/dt = 1/2 * 4 = 2 in^3/second</p>

3

<p>Oh wow thanks a lot! I don’t know what happened to me when I did this but I kept thinking that the area meant the area of one side…hahahaha…</p>

<p>So can you help me with question 2 now? How can I get the slope at x=1 when i don’t know what the y value is at x=1?</p>

<p>Thanks a lot!</p>

4

<p>It’s a function - how do you find y when you are given x?</p>

<p>You can then find dy/dx by implicit differentiation (aka the chain rule)</p>

5

<ol>
<li>What is the approximate slope of the tangent to the curve x^3+y^3=xy at x = 1? </li>
</ol>

<p>Implicit differentiation</p>

<p>Start by taking the derivative of the left side, and then chain rule for the right
3x^2 + (3y^2)dy/dx = y + x(dy/dx)</p>

<p>Get the dy/dx ‘s to the same side and solve for it. That is the derivative/slope so thats where you’ll plug in the value x=1. You will also need a y-value so plug x=1 into the original function to get the y-value, plug that into f’(x)</p>