The radius of a circular plate is increasing at the rate of 0.01 cm / sec. When the radius is 12 cm, the rate at which the area increases is (in square cm / sec )

Question

The radius of a circular plate is increasing at the rate of 0.01 cm / sec. When the radius is 12 cm, the rate at which the area increases is (in square cm / sec ) (A) 60 π (B) 24 π (C) 1.2 π (D) 0.24 π

Tardigrade Admin · Accepted Answer

(d r/d t)=0.01 cm / s, when r=12 cm, then (d A/d t)= ? A=π r2 ⇒ (d A/d t)=2 π r (d r/d t) ⇒ (d A/d t)=2 π(12) ⋅(0.01) ⇒ (24/100) π ⇒ 0.24 π

Q. The radius of a circular plate is increasing at the rate of $0.01 c m / sec$ . When the radius is $12 c m$ , the rate at which the area increases is (in square $c m / sec$ )

$60 π$

$24 π$

$1.2 π$

$0.24 π$

$\frac{d r}{d t} = 0.01 c m / s$ , when $r = 12 c m$ , then $\frac{d A}{d t} =$ ?
$A = π r^{2}$
$\Rightarrow \frac{d A}{d t} = 2 π r \frac{d r}{d t}$
$\Rightarrow \frac{d A}{d t} = 2 π (12) \cdot (0.01)$
$\Rightarrow \frac{24}{100} π$
$\Rightarrow 0.24 π$

Q. The radius of a circular plate is increasing at the rate of 0.01cm/sec. When the radius is 12cm, the rate at which the area increases is (in square cm/sec )

60π

24π

1.2π

0.24π