Question

The radius of a cylinder is increasing at the rate of 5 cm/min so that its volume is constant. When its radius is 5 cm and height is 3 cm, then the rate of decreasing of its height is
A) 6 cm/min
B) 3 cm/min
C) 4cm/min
D) 5 cm/min
E) 2 cm/min

Answer 1

Hint : We will take ‘v’ as the volume of the cylinder. Then we will write the formula for the volume of the cylinder then we will differentiate it with respect to t and equalize it with zero. There we will get dh/dt the required answer for the given solution.
Formula used:
The volume of the cylinder with height ‘h’ and radius ‘r’ is $V = \pi {r^2}h$

Complete step-by-step answer :
Let consider V as the volume of the cylinder.
In the question we are given that radius of a cylinder is increasing at the rate of 5 cm/min so we can represent it in $\dfrac{{dr}}{{dt}} = 5$ ,
r=5 cm and h=3 cm
We have the formula of volume that is
$V = \pi {r^2}h$
Differentiating it with respect to ‘t’, we get
$\dfrac{{dv}}{{dt}} = 2\pi rh\dfrac{{dr}}{{dt}} + \pi {r^2}\dfrac{{dh}}{{dt}}$
Now we have the volume is constant whatever may be the rate of increase of radius.
Since, V is constant $\dfrac{{dv}}{{dt}} = 0$
$\Rightarrow 2\pi rh\dfrac{{dr}}{{dt}} + \pi {r^2}\dfrac{{dh}}{{dt}} = 0$
Now we can take out π and r from this and we will take the rest to the other side of the equation. So now we have
$\Rightarrow 2h\dfrac{{dr}}{{dt}} = - r\dfrac{{dh}}{{dt}}$
In the question we are given that $\dfrac{{dr}}{{dt}} = 5$ and ‘h’ is 3cm and ‘r’ is 5 cmso we will substitute this value in the given equation and hence we have
$\Rightarrow 2 \times 3 \times 5 = - 5\dfrac{{dh}}{{dt}}$
$\Rightarrow \dfrac{{dh}}{{dt}} = - 6$
Hence the rate of decrease of height, $- \dfrac{{dh}}{{dt}} = 6$ cm/min.
So, the correct answer is “Option A”.

Note : If the final answer has negative sign then it is decreasing and if it has positive sign then it is increasing. So do not confuse about getting the rate as a negative sign and choose the options wisely.