Question

radius of a circle increase at the rate of 2cm/sec at what rate area increase when a radius is 10 cm​

Answer 1

Cσrrєcт Qυєѕтiσи —

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The radius of a circle is increasing uniformally at the rate of 3 cm/sec . Find the rate at which the area of circle is increasing when the radius is 10 cm.

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Let :

• r be the radius of circle .
• A be the area of circle .

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Giνєи :

• Radius of circle is increasing at the rate of 3 cm/sec .
• $\sf{\longmapsto \dfrac{dr}{dt}=3cm^s}$

✪ We need to find the rtae of change of area of circle w.r.t time , when r = 10cm. i.e,

$\qquad\rm{\longrightarrow \dfrac{dA}{dt}}$ , when r = 10 cm

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✪ We know that :

$\qquad{\pmb{\rm{Area_{(circle)}=\pi r^2 }}}$

✪ Differentiating w.r.t time :

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

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

$\qquad\sf{\longrightarrow \dfrac{dA}{dt}= \pi\dfrac{d(r^2)}{dt}\times \dfrac{dr}{dr}}$

$\qquad\sf{\longrightarrow \dfrac{dA}{dt}=\pi \dfrac{d(r^2)}{dr}\times \dfrac{dr}{dt} }$

$\qquad\sf{\longrightarrow \dfrac{dA}{dt}= \pi .2r.\dfrac{dr}{dt}}$

$\qquad\sf{\longrightarrow \dfrac{dA}{dt}=2\pi r .\dfrac{dr}{dt} }$

$\qquad\sf{\longrightarrow \dfrac{dA}{dt}=2\pi r .3 }$

$\qquad\sf{\longrightarrow \dfrac{dA}{dt}= 6\pi r}$

➻ Where r = 10 cm

$\qquad\sf{\longrightarrow \dfrac{dA}{dt}\bigg |_{r=10}=6\times \pi \times 10 }$

$\qquad\sf{\longrightarrow \dfrac{dA}{dt}\bigg | _{r=10} = 60\pi }$

➻ Since , area is in cm ² , and time is in sec :

$\qquad{\pmb{\rm{\longrightarrow \dfrac{dA}{dt}=60\pi cm^2 /sec }}}$

❝ Hence , area is increasing at the rate of 60π cm²/ sec , when r is 10cm ❞

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Answer 2

Answer:

ABC revolt circle of rid of circle of

Step-by-step explanation:

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