Solving a Calculus Question: Cone with Radius 5 ft & Height 10 ft

[hoedown_j]

Calculus question...

I had this on a test yesterday, and for some reason it seems like it was way too simple.

A cone with a radius of 5 ft and height of 10 ft is being filled with water at a rate of 3 ft^3/min. Find the rate at which the height of the water is changing when the height is 7 ft.

So I have V=1/3 B * H, with the base being 25 pi, and dv/dt being 3 ft^3/min.

dv/dt=1/3(25 * pi * dh/dt)
3=1/3(25 * pi * dh/dt)
9=25 * pi * dh/dt
9/(25pi)=dh/dt
dh/dt=0.1146 ft/min
Is this correct? This is what I got the first time I did the problem, but I remember writing down my answer as 1.14 ft/min because that was the answer I got the second time I did it. It seems too simple, especially because this was worth twice as many points as any other problems on the test.

 

[courtrigrad]

The volume of a cone is:

v = 1/3 * pi * x^2 * y.
v = 1/12 * pi * y^3

dv/dt = 3 ft^3 / min.
We want to find dy/dt when y = 7.

dv/dt = 1/4* pi * y^2 * dy/dt

Hence dy/dt = (4/ pi*y^2 ) * dv/dt

When dv/dt = 3 and y = 7 gives

12/ 49pi whuch is approx 0.08 ft^2 / min

This may have been worth twice as much as the other problems because you had to use similar triangles to obtain x in terms of y. In this case,

x / y = 1/2

x = 1/2y.

 

[dextercioby]

V(t)=\frac{1}{3}\pi r^{2}(t)h(t) (1)
Diff.wrt to time:

\frac{dV(t)}{dt}=\frac{\pi}{3}[2r(t)\frac{dr(t)}{dt}h(t)+r^{2}(t)\frac{dh(t)}{dt}] (2)

At the moment "t":
h=7;\tan\alpha=1/2 \Rightarrow r=7\tan\alpha=3.5(3)
r=h\tan\alpha \Rightarrow \frac{dr(t)}{dt}=\frac{1}{2}\frac{dh(t)}{dt}(4)

Plug (3) and (4) in (2) and find
\frac{dV(t)}{dt}=\frac{\pi}{3}[2\cdot (3.5)\cdot 7\cdot\frac{1}{2}\frac{dh(t)}{dt}+(3.5)^{2}\frac{dh(t)}{dt}]\Rightarrow \frac{dh(t)}{dt}=\frac{3\frac{dV(t)}{dt}}{\pi[7\cdot (3.5)+(3.5)^{2}]}
=\frac{3\cdot 3}{\pi(24.5+12.25)}=\frac{9}{\pi\cdot 36.75}\sim 0.08ft.s^{-1} (5)

Daniel.