Question

The diameter of a sphere is measured to be 40cm. If an error of 0.02 cm is made in it, then find the approximate errors in volume and surface area of the sphere.

Answer 1

Hint: To solve these types of questions, we have to take use of differentials. To find out the approximate error in volume or in any quantity, we should differentiate that quantity with respect to the quantity that already has error. In this case, we should differentiate volume with respect to radius.

Complete step-by-step solution:

Given, the diameter of a sphere is 40 cm

So, the radius of the sphere is equal to the half of the diameter as

$ \Rightarrow radius = \dfrac{(\text{diameter})}{2}$ 

$ \Rightarrow \dfrac{40}{2} = 20{\text{ cm}}$ 

Now, the error in measuring the diameter is 0.02 cm.

So, the error in radius will be equal to 0.01 cm and it is represented as $\Delta {\text{r}}$ 

Now, we have to find the approximate errors in volume as well as in the surface area of the sphere which will be represented as $\Delta {\text{v}}$ and $\Delta {\text{s}}$ respectively.

To find the approximate error in volume, we have to differentiate V with respect to r and multiply it with $\Delta {\text{r}}$ as

$\Delta {\text{v }} =  \left(\dfrac{d}{{dr}}v\right)\times \Delta {\text{r}}$ 

The volume of sphere is given by $\dfrac{4}{3}\pi {r^3}$ 

Now after differentiating and putting values, we get

$\Delta {\text{v  =  }}\dfrac{4}{3}\pi (3{r^2}) \times (0.01)$ 

$ = 50.24{\text{ }}c{m^3}$ 

Similarly, the surface area of sphere is given by $4\pi {r^2}$ 

To find the approximate error in surface area, we have to differentiate S with respect to r and multiply with $\Delta {\text{r}}$ as

$\Delta {\text{s}}  =  \left(\dfrac{{ds}}{{dr}}\right)\Delta r$ 

After differentiating and assigning values, we get

$   \Rightarrow \Delta {{s  =  (4\times2\times}}\pi {\text{r)}}\Delta {\text{r}} $

$   \Rightarrow \Delta {\text{s  =  (8}}\pi {{\times20)(0}}{\text{.01)}} $

$   \Rightarrow \Delta {\text{s  =  (1}}{\text{.6}}\pi ) $

$   = 5.024{\text{ c}}{{\text{m}}^2} $ 

$\therefore$ The approximate error in calculating volume is $50.24cm^3$ and approximate error in calculating surface area is $5.024 cm^2$

Note: We use differentials to solve these types of questions because differentials provide us with a way of estimating the amount of function changes when a small change occurs in any input value. Here a doubt may arise in mind as to why we are multiplying our differentials with the error. It is because we have to maintain the units on both sides as well.