Question

The radius of the base of a certain cone is increasing at the rate of 3 cm per minute and the
altitude is decreasing at the rate of 4 cm per minute. Find the rate of change of total surface of
the cone when the radius is 7 cm and the altitude is 24 cm.​

Answer 1

Given :-

The increasing rate of change of radius of a cone = 3 cm/min

The decreasing rate of change of height of a cone = -4 cm/min

Radius of the cone = 7 cm

Altitude of the cone = 24 cm

To Find :-

The rate of change of total surface of the cone we need to differentiate.

Solution :-

We know that,

• r = Radius
• h = Height
• dh/dt = The rate of change of height with respective to time.
• dr/dt = Differentiating r in respective with t

Given that,

Radius (r) = 7 cm

Altitude (h) = 24 cm

(dr/dt) = 3 cm/min

(dh/dt) = -4 cm/min

Let 'r' be the radius, 'h' be the height and 's' be the total surface area

Now,

$\sf s= \pi r^{2}+\pi rl=\pi r^{2}+\pi r\sqrt{r^{2}+h^{2}}$

$\sf \dfrac{ds}{dt} =2 \pi r \dfrac{dr}{dt} + \pi \bigg( \dfrac{d}{dt} \sqrt{r^{2}+h^{2}} +\sqrt{r^{2}+h^{2}} \dfrac{dr}{dt} \bigg)$

$\sf \dfrac{ds}{dt} =2 \pi r(3)+ \pi \bigg[ r\dfrac{1}{2\sqrt{r^{2}+h^{2}} } \bigg(2r\dfrac{dr}{dt} +2h \dfrac{dh}{dt} \bigg) +\sqrt{r^{2}+h^{2}} \dfrac{dr}{dt} \bigg]$

$\sf \dfrac{ds}{dt} =6 \pi r+ \pi \bigg[ \dfrac{r}{\sqrt{r^{2}+h^{2}} } (r(3)+h(-4))+\sqrt{r^{2}+h^{2}} (3) \bigg]$

$\sf \therefore \dfrac{dx}{dt} \left \{ {{r=7} \atop {h=24}} \right =6 \pi (7)+ \bigg[ \dfrac{7}{\sqrt{(7)^{2}+(24)^{2}} } (7(3)+24+(-4))+\sqrt{(7)^{2}+(24)^{2}} (3) \bigg]$

$\sf 42 \pi + \pi \bigg[ \dfrac{7}{\sqrt{625} }(-7)+\sqrt{625} (3) \bigg]$

$\sf 42 \pi+\pi \bigg[ \dfrac{7}{25} (-75)+25(3) \bigg]$

$\sf 42 \pi+\pi (-21+75)$

$\sf 42 \pi +54 \pi$

$\sf =96 \pi \ sq \ cm/min$