Question

Find the rate of change of lateral surface area of a cone w.r:t. to its radius, when the height is kept constant.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$\pi \sqrt{h^{2}+r^{2} }+\frac{\pi r^{2} }{\sqrt{h^{2}+r^{2}  } }$

Step-by-step explanation:

Let us assume that we have a cone with a radius of the base circle r and height h.

Then the lateral surface area of the cone is given by A=πr√(h²+ r²).

So, the rate of change of lateral surface area(A) of the cone with respect to its radius(r), when the height(h) is kept constant, will be

$\frac{dA}{dr}=\frac{d}{dr}(\pi r\sqrt{h^{2}+r^{2}  }  )$

We have to use the formula of differentiation  

$\frac{d}{dx} (u.v)= v.\frac{du}{dx}+u.\frac{dv}{dx}$

⇒ $\frac{dA}{dr}= \pi \sqrt{h^{2}+r^{2} }+\frac{\pi r(2r)}{2\sqrt{h^{2}+r^{2}  } }$

⇒$\frac{dA}{dr}=\pi \sqrt{h^{2}+r^{2} }+\frac{\pi r^{2} }{\sqrt{h^{2}+r^{2}  } }$

(Answer)