Calculus- Surface area + Rate of change of Volume?

Question

A right circular cone and a hemisphere have the same base, and the cone is inscribed in the hemisphere. The figure is expanding in such a way that the combined surface area of the hemisphere and its base are increasing at a constant rate of 18 square inches per second. At what rate is the volume of the cone changing at the instant when the radius of the common base is 4 inches?

The surface area of a sphere of radius r is
S=4*pi*r^2
and the volume of a right circular cone of height h and radius r is
V=(1/3)*pi*(r^2)*h


I have taken the equation of the sphere and divided it by 2 for the equation of the Surface Area of a hemisphere then I found the derivative of that which makes the rate of change of the SA to be 8*pi.
Then I took that and divided by 18(in^2)/second
which comes out as (4*pi/9) seconds.
Here is where I am stuck and I do not know if the work I have done is even remotely close to my final goal rate of change of volume of the cone at 4 radius h.

Help Plz

Answer 1

OK, well you started out right, but then you veered off. So, the surface area (SA) of a sphere is 4*pi*r^2. We divide by two and get sufrace area of hemishpere is 2*pi*r^2. Remmeber the problem says that the COMBINED surface area of the hemisphere AND base are increasing at.... So now we need to add the surface are of the base. It is a circle, so the area is pi*r^2. The combined SA is 2*pi*r^2 + pi*r^2 = 3*pi*r^2. The rate of change for SA is a constant and equals 18 in^2 per second. We have the equation for SA = 3*pi*r^2. The radius changes as a function of time r = f(t) but we don't know this function yet. We do know that the derivative of SA with respect to time is 18. We can use the chain rule to tell us that dSA/dr * dr/dt = dSA/dt.
We can find dSA/dr; that's easy (6*pi*r)

We also know that dSA/dt is 18 in^2/sec. Now we can solve the chain rule equation to get dr/dt = 18in^2/sec / (6*pi*r)
The question asks about the instant when r = 4 inches. So we will plug in r=4 inches to find that dr/dt (when r=4) is 18 in^2/(sec*6*pi*4 inches) = 3in/(sec*4*pi) Now we need to use the chain rule again. You have an equation for V. Find dV/dr. We just solved for dr/dt. You know that dV/dr*dr/dt will give you dV/dt which is the answer you're looking for.

Answer 2

The first thing to notice is that, if the cone is inscribed *in* the hemisphere, then it radius and height are the same.

Next, the problam says that "the combined surface area of the hemisphere and its base" are increasing at 18 in²/s. So you need an expression for the area of the circular base of the hemisphere (πr² sound 'bout right ☺?) The sum of these two is the total area of the hemisphere and it's base.

Next, you need to find the rate of change of r with respect to t (that'd be the derivative, dr/dt) when dA/dt = 18 (take the derivative of the equation you got for surface area with respect to t) and solve it for dr/dt when r = 4

Now take the derivative of the expression for the volume of the cone (with respect to time) and evaluate dV/dt with r = 4 and dr/dt being whatever it was that you got above.

That's it ☺


Doug

Answer 3

The total surface area of the hemisphere plus the base is 3*pi*r^2. Define the derivative of r with respect to time to be v. Then the derivative of the surface area with respect to time is 6*pi*r*v. We then have:

6*pi*(4)*v=18 (The units will turn out to be consistent, so I'm not writing them down.)

We therefore have v.

For the cone, note that h=r (Why?).

Therefore, V=(1/3)*pi*(r^3). Let the derivative of V (don't confuse this with v, above) with respect to time be W. Then:

W=pi*r^2*v.

Now just replace r with 4 and v with what you find from above, and you get your answer, i.e., you get W.

Answer 4

Let the radius of base be r, so the height of the cone is also r

Surface Area of hemisphere and circular base

S= 2*pi* r^2 + pi* r^2 = 3*pi* r^2

dS/dt = 3* pi* 2* r* dr/dt = 6*pi*r*dr/dt

dS/dt is given to be 18, therefore

6*pi*r*dr/dt = 18,

dr/dt = 3/(pi*r)

Now Volume of cone (V)

V= 1/3* pi * r^3

dV/dt = pi * r^2* dr/dt = pi * r^2*3/(pi*r)= 3r

dV/dt at r=4 is 12, that's the answer.

Answer 5

slowly differciate the known variable