# The surface area of a circular cone with radius r and height h is A=\frac{1}{3}\pi...

## Question:

The surface area of a circular cone with radius {eq}r{/eq} and height {eq}h{/eq} is {eq}A=\frac{1}{3}\pi r\sqrt{r^2+h^2}{/eq}. Suppose that r decreases from 12 cm to 11.5 cm, while {eq}h=16{/eq} cm remains constant. Then find the approximate change in the area of the cone as given by the differentials.

## Insert context header here: Change in surface Area of cone

Surface area of a cone is given by

{eq}\displaystyle A=\frac{1}{3}\pi r\sqrt{r^2+h^2} {/eq}

{eq}\displaystyle \frac{d(\ln x)}{dx} = \frac{1}{x} {/eq}

Given

radius of the cone = r

height of the cone = h = 16 cm

Initial radius of the cone = 12 cm

Final radius of the cone = 11.5 cm

{eq}\displaystyle dr = (12 - 11.5)\\ dr = 0.5\\ \frac{dr}{r} = \frac{0.5}{12}\\ \frac{dr}{r} = \frac{1}{24}\\ {/eq}

Initial area of the cone

{eq}\displaystyle A = \frac{1}{3}\pi r\sqrt{r^2+h^2}\\ A = \frac{1}{3}\pi (12)\sqrt((12)^2 + (16)^2)\\ A = 80\pi {/eq}

Area of the cone

{eq}\displaystyle A = \frac{1}{3}\pi r\sqrt{r^2+h^2}\\ {/eq}

Apply ln on both sides

{eq}\displaystyle \ln A = \ln(\frac{1}{3}\pi r\sqrt{r^2+h^2})\\ \ln A = \ln(\frac{1}{3}\pi) + \ln r +\frac{1}{2} \ln(r^2 + h^2)\\ {/eq}

differentiate

{eq}\displaystyle \begin{align} \frac{dA}{A} &= \frac{dr}{r} +\frac{1}{2} \frac{2dr}{(r^2 + h^2)}\\ \frac{dA}{A} &= \frac{dr}{r} + \frac{dr}{(r^2 + h^2)}\\ \frac{dA}{80\pi} &= \frac{0.5}{12} + \frac{dr}{((12)^2 + (16)^2)}\\ \frac{dA}{80\pi} &= \frac{1}{24} + \frac{1}{800}\\ dA &= 10.7806667 \end{align} {/eq}