If the radius of a sphere is increased by $10\%$, by what percentage is its volume increased? 
Question: If the radius of a sphere is increased by $10\%$, by what percentage is its volume increased? Use Calculus.
Answer: It increases by approximately $33.1\%$.


I did the above question using calculus but my answer came out to be 30%. Here's the solution. What is wrong in it?

$V = \frac{4}{3} πr^3$
$dV = \frac 4 3  π \cdot 3  r^2 dr$
$dV  = 4 π r^2 dr$
since
$dr = \frac{10}{100} r$
therefore
$dV = 4  π  r^2 \cdot \frac{10}{100}  r$
$dV = \frac 4 {10} π r^3$
now
$dV/V = [ \frac 4{10}  πr^3] / [ \frac 4 3  πr^3]$
$dV/V = 3/10$
and
$100dV/V = \frac 3{10} \times 100 = 30\%  $

The picture below shows you what you did wrong. You used the gradient function at a point $r_1$ and extrapolated that tangent line to approximate the value of $\frac 4 3 \pi r^3$ at a value $r_2$ which is 10% bigger than $r_1$. In the diagram below, the red line is the tangent line to $V$, which increases from $1$ to $1.3$ (as shown by the light green line) as $r$ increases from $r_1$ to $r_2=1.1\times r_1$. On the other hand, the value of the volume function (brown) at $r_2$ is $1.331$, as shown by the dotted blue line. There is a discrepancy between the two values because point $A$ is not in the same position as point $B$

The best way to solve your problem is simply to calculate the difference in the values of $V$ (from point $(0.62,1)$ to point $B$ in the diagram above): $$\frac{V(r_2)-V(r_1)}{V(r_1)}=\frac{\frac 4 3\pi (1.1r)^3 -\frac 4 3 \pi r^3}{\frac 4 3\pi r^3}=\frac{1.1^3-1}1=0.331=33.1\%$$ where $r_2=1.1\times r_1$.
