# Cone shaped related rates of change question

A container is in the shape of a cone of semi-vertical angle $30^\circ$, with it's vertex downwards.

Liquid flows into the container at ${{\sqrt {3\pi } } \over 4}{\rm{ }}c{m^{^3}}/s$

At the instant when the radius of the circular surface of the liquid is 5 cm, find the rate of increase of:

(a) The radius of the circular surface of the liquid

(b) The area of the circular surface of the liquid

My attempt:

${{dV} \over {dt}} = {{\sqrt {3\pi } } \over 4}$

(A) I need to find the rate at which the radius increases as "h" increases, so I have to find ${{dr} \over {dt}}$.

The equation for the volume of a cone is:

$V = {1 \over 3}\pi {r^2}h$

I now must form a function in terms of r for h. As we are asked to a compute when the radius is 5 we can form an equation using similar triangles, so:

\eqalign{ & {h \over x} = {r \over 5} \cr & x: \cr & \tan 30^\circ = {5 \over x} \cr & x = {5 \over {\tan 30^\circ }} \cr & x = 5\sqrt 3 \cr & so: \cr & h = r\sqrt 3 \cr}

\eqalign{ & V = {1 \over 3}\pi {r^2}(r\sqrt 3 ) \cr & V = \pi {r^3}{{\sqrt 3 } \over 3} \cr & {{dV} \over {dr}} = \pi {r^2}\sqrt 3 \cr & {{dr} \over {dt}} = {{dV} \over {dt}} \times {{dr} \over {dV}} \cr & {{dr} \over {dt}} = {{\sqrt {3\pi } } \over 4} \times {1 \over {\pi {r^2}\sqrt 3 }} \cr & {{dr} \over {dt}} = {{\sqrt {3\pi } } \over {\pi {r^2}4\sqrt 3 }} = {{\sqrt \pi } \over {\pi {r^2}4}} \cr}

\eqalign{ & r = 5: \cr & {{\sqrt \pi } \over {4(25)\pi }} = {{\sqrt \pi } \over {100\pi }} = 0.005641... \cr}

For part (A) the answer is stated as 0.01 cm/s, nowhere in the question have I been asked to round my answer, have I obtained the correct answer? I just want to make sure..

Part (b)

Part (B) requires that I calculate the rate of increase of the circular area of the liquid, so essentially ${{dA} \over {dt}}$.

Area of a circle is: $A = \pi {r^2}$

\eqalign{ & A = \pi {r^2} \cr & {{dA} \over {dr}} = 2\pi r \cr & {{dA} \over {dt}} = {{dr} \over {dt}} \times {{dA} \over {dr}} \cr & {{dA} \over {dt}} = {{\sqrt \pi } \over {4\pi {r^2}}} \times 2\pi r \cr & {{dA} \over {dt}} = {{\sqrt \pi } \over {2r}} \cr & r = 5: \cr & {{dA} \over {dt}} = 0.1\sqrt \pi {\rm{ c}}{{\rm{m}}^2}/s \cr}

However the answer in the book for b is ${{dA} \over {dt}} = 0.1\pi {\rm{ c}}{{\rm{m}}^2}/s$ (pi is not square rooted), Where have I gone wrong?

Furthermore I'd love it if any answerers could suggest how I could improve on how I've done things, and any tricks/tips that would make things easier for myself in the future.

Thank you.

• If you are going to use $x$, you should define it. In fact, you don't need to find $h$, but if you want to, $\frac rh=\tan 30^\circ=\frac {\sqrt 3} 3$, so $h=5\sqrt 3$, which you got. – Ross Millikan May 2 '13 at 22:37
• @RossMillikan, Oh okay, I'll bear that in mind, thank you! – seeker May 2 '13 at 22:54

"Liquid flows into the container at: $\;\displaystyle \dfrac{\sqrt{3}\,\pi}{4} = {{(\sqrt {3})\,\pi} \over 4}{\rm{ }}\text{ cm$^3$per sec}$"
This would explain both discrepancies, (between both the solutions you obtained, and the solutions of the text), since your work seems to be fine, and was clearly done carefully. Simply changing your final evaluations using $\pi$ instead of $\sqrt{\pi}$ will yield the solutions you are given.