related rates problem: right circular cone, hemisphere share

[alakaboom1]

This problem has me extremely confused, would really appreciate some help..

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 is increasing at a constant rate of 18 square inches per second. At what rate is the volume of the cone changing at the instan when the radius of the common base is 4 inches?

As i said, im extremely confused and have nothing....Thanks in anyone can help!!

 

[Deleted member 4993]

Re: Confusing related rates problem

This problem has me extremely confused, would really appreciate some help..

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 is 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?

As i said, im extremely confused and have nothing....Thanks in anyone can help!!

First draw a sketch of the situation.

The radius of common base = r

height of the hemisphere and the cone = r

what is the surface area of the hemisphere (including the base) = A[sub:fok6rxl6]s[/sub:fok6rxl6] = 2 * (pi) r^2 + (pi)r^2 = 3(pi)r^2

What is the rate of change of surface area = dA[sub:fok6rxl6]s[/sub:fok6rxl6]/dt = dA[sub:fok6rxl6]s[/sub:fok6rxl6]/dr * dr/dt = 6(pi)r * dr/dt = 18

dr/dt = 3/[(pi)r]

What is the Volume of the cone = V[sub:fok6rxl6]c[/sub:fok6rxl6] = ??

What is the rate of change of the Volume = dV[sub:fok6rxl6]c[/sub:fok6rxl6]/dt = dV[sub:fok6rxl6]c[/sub:fok6rxl6]/dr * dr/dt

and continue.....

 

[BigGlenntheHeavy]

Re: Confusing related rates problem

Let A = total area of hemisphere = 3?r²

Volume of cone = ?r³/3

dA/dt = 6?r(dr/dt). When dA/dt = 18in²/sec and r = 4in., then

18in²/sec = 24?in(dr/dt), dr/dt = 3in/4?sec.

Now dV/dt = ?r²(dr/dt) = [16?(in)²(3in)]/4?sec. = 12(in)³/sec.

Got it?

 

[galactus]

This site provides nice tutorials on related rates. At least this one was a little different than the same old cliche problems that have been done to death.

http://tutorial.math.lamar.edu/Classes/ ... Rates.aspx