“An ice cream cone with a radius of 2 inches at the top and 6 inches tall has a little hole in the bottom. The melted ice cream is one inch from the top of the cone. What is the rate that the volume of the melted ice cream in the cone is decreasing with respect to the height of the ice cream in the cone?”
The answer is dV/dx = 25π/9 at x = 5
I know the volume of a cone = 1/3(πr^2)h, but other than that I’m stuck :(
On
Note the radius $r$ and the height $h$ is related via $\frac rh = \frac 26 = \frac 13$. The volume is then,
$$V= \frac13 \pi r^2 h = \frac1{27}\pi h^3$$
The rate that the volume decreases with respect to the height when the melting ice cream is one inch from the top, or $h=5$, is
$$\frac{dV}{dh} = \frac1{9}\pi h^2= \frac{25\pi}{9}$$
Hint: We can write the radius of the cone in terms of the height using similar triangles in a cone. Since the ratio of the height to the radius is six to two, we know $r = h/3$ at all times. So we can rewrite the volume as $$V = \frac{1}{9}\cdot \frac{\pi}{3}h^3 = \frac{\pi}{27}h^3 \implies \frac{dV}{dh} = \frac{\pi}{9}h^2$$ For some arbitrary height $h$.