Question

the rate of change of the area of a cricle with respect to its radius r at r=5 is​

Answer 1

Given :

• radius of circle = 5 unit

To find :

• Rate of change of area ( dA/dr) of the circle with respect to radius.

Formula used :

• A = πr²

where :

• A = area if circle
• r = radius of circle
• π = 22/7

Solution :

Now , radius of circle = 5 unit.

$\implies \: A = \pi {r}^{2}$

Now differentiating both sides :-

$\implies \: \dfrac{d(A)}{dr} = \dfrac{d(\pi{r}^{2} )}{dx}$

$\implies \: \dfrac{d(A)}{dr} = \pi \: \dfrac{d({r}^{2} )}{dx}$

$\implies \: \dfrac{d(A)}{dr} = \pi \: 2r \: \dfrac{d({r})}{dr}$

Now put r = 5 and cancel out (dr) from numerator and denominator :-

$\implies \: \dfrac{d(A)}{dr} = \pi \times \: 2 \times 5$

$\implies \: \dfrac{d(A)}{dr} = \pi \times \: 10$

$\implies \: \dfrac{d(A)}{dr} = 10 \pi$

Put π = 22/7

$\implies \: \dfrac{d(A)}{dr} = \dfrac{22}{7} \times \: 10$

$\implies \: \dfrac{d(A)}{dr} = \dfrac{220}{7}$

Answer :

$\dfrac{d(A)}{dr} = \dfrac{220}{7} \: or \: 10 \pi$

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$\large\bf\blue{Additional-Information}$

Area of circle = πr²

Circumference of circle = 2πr

Area of square = (side)²

perimeter of square = 4× side

area of Rhombus = 1/2(diagonal 1)(diagonal 2)

Area of rectangle = length × breadth

Perimeter of rectangle = 2( length + breadth)

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d(x^n)/dx = n x^(n-1)

d(e^x)/dx = e^x

d(logx)/dx = 1/x

d(sinx)/dx = cosx

d(cos x)/dx = -sin x

d(cosec x)/dx = -cot x cosec x

d(tan x)/dx = sec²x

d(sec x)/dx = secx tanx

d(cot x)/dx = - cosec² x

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