Can someone help me with a few calculus problems?

**calvinandhobbes** · August 12th 2010, 09:40 PM

1.The volume of a cube is increasing at 120 cubic centimeters per minute at the instant when its edges are 8 cm. At what rate are the edge lengths increasing at that instant?

I know the answer is 5/8 cm per min but i don't know how. If anybody can go through the steps, that would be appreciated.

2. Express the surface area S of a sphere as a function of its volume V, then find dS/dV.

3.What is the slope of line that goes through the origin and intersects the curve y=2^x tangentially, that is in one point only? Slope of line and point of intersection? Can you explain how you got the answer? thx

4.Write an equation for the line tangent to the parabola x= 8y - y^2 at the point (15,3)?

**Prove It** · August 12th 2010, 09:51 PM

2. $S = 4\pi r^2$

$V = \frac{4\pi r^3}{3}$

$3V = 4\pi r^3$

$\frac{3V}{4\pi} = r^3$

$\left(\frac{3V}{4\pi}\right)^{\frac{1}{3}} = r$

$\left(\frac{3V}{4\pi}\right)^{\frac{2}{3}} = r^2$

$\frac{(3V)^{\frac{2}{3}}}{(4\pi)^{\frac{2}{3}}} = r^2$

$\frac{4\pi (3V)^{\frac{2}{3}}}{(4\pi)^{\frac{2}{3}}} = 4\pi r^2$

$(4\pi)^{\frac{1}{3}}(3V)^{\frac{2}{3}} = 4\pi r^2$

$[4\pi(3V)^2]^{\frac{1}{3}} = 4\pi r^2$

$[4\pi(9V^2)]^{\frac{1}{3}} = 4\pi r^2$

$(36\pi V^2)^{\frac{1}{3}} = 4\pi r^2$

$S = (36\pi V^2)^{\frac{1}{3}}$ .

Now you will need to use the Chain Rule to find $\frac{dS}{dV}$ .

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