Answer to Question #156657 in Calculus for Phyroe

Question

The base diameter and altitude of a right circular cone are observed at a certain instant to be 10 and 20 inches, respectively. If the lateral area is constant and the base diameter is increasing at a rate of 1 inch per minute, find the rate at which the altitude is decreasing.

Answer 1

$Solution: Let ~r ~be~ diameter~ and~ h~ be~ the~ altitude~ of~ the~ right~ circular ~~cone. \\Then~ lateral ~area ~of~ right ~circular~ cone~ is ~=A=πr\sqrt{h^2+r​​​​​​^r}$

$Differentiate ~~implicitly~ with ~respect~ to ~t \\\frac{d}{dt}(A)=\frac{d}{dt}(πr\sqrt{h^2+r​​​​​​^r})$

$\Rightarrow \frac{dA}{dt}=\frac{\pi}{2}[\frac{r^2}{2}\frac{1}{2}\sqrt{(\frac{r^2}{2}+h^2)(r\frac{dr}{dt}+2h\frac{dh}{dt})}+\sqrt{\frac{r^2}{2}+h^2r\frac{dr}{dt}}] ...................................(I)$

$Given~that~\frac{dA}{dt}=0. Also ~r=10~inches~and~h~=20~inches~and~ \frac{dr}{dt}=0$

$Put~ all ~values ~in ~equation~ (I),then ~we~ get, \\351.87+74.09\frac{dh}{dt}=0 \\74.09\frac{dh}{dt}=-351.87 \\\frac{dh}{dt}=-\frac{351.87}{74.09} \\\frac{dh}{dt}=-4.75 \\Negative ~sign~ indicates~ that~ altitude ~is ~decreasing. \\Thus,~the~ rate ~at~ which ~altitude ~is~ decreasing ~be ~4.75~ per ~minute.$