Question

The radius of a circular plate increases at the rate of 0.1 cm/sec. At what rate does the area increase when the radius of plate is
11 cm.​

Answer 1

Answer:

Rate of increase of area = 6.908 cm²/s

Explanation:

Given:

• The radius of a circular plate increases at the rate of 0.1 cm/s

To Find:

• The rate of increase in area when the radius of the plate is 11 cm

Solution:

We know that,

Area of a circular plate = π r²

where r is the radius of the plate.

By given the radius of the plate is increasing at the rate of 0.1 cm/s.

That is,

$\sf \dfrac{dr}{dt} =0.1\:cm/s---(1)$

where t is the time taken

Now we have to find the rate of increase of area, $\sf \dfrac{dA}{dt}$

Substituting the value of area we get,

$\sf \dfrac{dA}{dt} =\dfrac{d}{dt} (\pi \: r^2)$

Differentiating we get,

$\implies \sf \pi\times 2r\: \dfrac{dr}{dt}$

Substitute the value of dr/dt from equation 1,

$\implies \sf 2\: \pi \: r\times 0.1$

$\implies \sf 0.2\: \pi\: r$

Now by given the radius of the plate is 11 cm.

Substituting it,

Rate of increase of area = 0.2 × 3.14 × 11

⇒ 6.908 cm²/s

Hence the rate of increase of area of the plate is 6.908 cm²/s.

Answer 2

Answer:

Given :-

• The radius of a circular plate increases at the rate of 0.1 cm/sec.

To Find :-

• At what rate does the area increase when the radius of plate is 11 cm.

Formula Used :-

$\clubsuit$ Area of Circle :

$\longmapsto \sf\boxed{\bold{\red{Area\: of\: Circle =\: πr^2}}}$

where,

• r = Radius

Solution :-

$\mapsto$ The radius of a circular plate increases at the rate of 0.1 cm/sec.

Then,

$\dashrightarrow \sf \dfrac{dr}{dt} =\: 0.1$

Now, we have to find the rate of increase in area :

$\implies \sf \dfrac{dA}{dt} =\: \dfrac{d}{dt}\bigg({\pi}r^2\bigg)$

$\implies \sf 2 \times {\pi} \times r \times \dfrac{dr}{dt}\: \bigg\lgroup \bold{By\: doing\: differentiating}\bigg \rgroup$

$\implies \sf 2{\pi}r \times \dfrac{dr}{dt}$

Here,

$\leadsto$ $\sf \dfrac{dr}{dt} =\: 0.1$

$\implies \sf 2{\pi}r \times 0.1$

$\implies \sf 2 \times 0.1\: {\pi}r$

$\implies \sf 2 \times \dfrac{1}{10}\: {\pi}r$

$\implies \sf \dfrac{2}{10}\: {\pi}r$

$\implies \sf\bold{\purple{0.2\: {\pi}r}}$

Now, we have to find the rate that does the area increase when the radius of plate is 11 cm :

Given :

• Radius = 11 cm

According to the question by using the formula we get,

$\longrightarrow$ Rate of increase of area :

$\Rightarrow \sf \dfrac{22}{7} \times 0.2 \times 11$

$\Rightarrow \sf \dfrac{22}{7} \times \dfrac{2}{10} \times 11$

$\Rightarrow \sf\dfrac{22 \times 2 \times 11}{7 \times 10}$

$\Rightarrow \sf \dfrac{\cancel{484}}{\cancel{70}}$

$\Rightarrow \sf\bold{\red{6.914\: cm^2/sec}}$

$\therefore$ The rate of increase when the radius of plate is 11 cm is 6.914 cm²/sec .