Question

using derivative, find the approximate percentage increase in the area of the circle if it's radius is increased by 2% please anyone help me to solve this..

Answer 1

i can help

Answer 2

\[ A = \pi r^2 \]
the change in A with respect to r i.e the derivative is
\[ \frac{dA}{dr} = 2 \pi r \frac{dr}{dr} = 2 \pi r \\ dA= 2 \pi r \ dr \]
if we replace the differential by the approximation
\[ \Delta A \approx 2 \pi r \Delta r \]

we want the fraction : change in A / A
\[ \frac{\Delta A}{A} = \frac{2 \pi r}{A} \Delta r\]

replace A with pi r^2
\[ \frac{\Delta A}{A}= \frac{2 \pi r}{\pi r^2} \Delta r = \frac{2}{r} \Delta r\]
multiply this fraction by 100 to get a percent change
\[ \% \text{ change in area} = \frac{200}{r} \Delta r \]

Answer 3

so far we have
\[ \text{percent change}= 200 \frac{\Delta r}{r} \]
we are told r changes by 2% so the fraction \( \frac{\Delta r}{r} = .02 \)
we get 200 * 2/100 = 4 (as a percent)

(I am thinking it would be less confusing to only work with fractions, i.e.
change = 2* Dr/r
then put in 2% for Dr/r ,and get 2*2% = 4%

but either way, the answer is about 4%

Answer 4

r u there