Short Questions (Question 20 – 22)

Posted on April 22, 2020 by user

Question 20:

The volume of water V cm³, in a container is given by

$V = \frac{1}{5} h^{3} + 7 h$ , where h cm is the height of the water in the container. Water is poured into the container at the rate of 15cm³s^-1. Find the rate of change of the height of water in cms^-1, at the instant when its height is 3cm.

Solution:

$\begin{array}{l} V = \frac{1}{5} h^{3} + 7 h \\ \frac{d V}{d h} = \frac{3}{5} h^{2} + 7 = \frac{3 h^{2} + 35}{5} \\ Given \frac{d V}{d t} = 15, h = 3 \\ Rate of change of the height of water = \frac{d h}{d t} \\ \frac{d h}{d t} = \frac{d h}{d V} \times \frac{d V}{d t} \leftarrow Chain rule \\ \frac{d h}{d t} = \frac{5}{3 h^{2} + 35} \times 15 \\ \frac{d h}{d t} = \frac{75}{62} {cms}^{- 1} \end{array}$

Question 21:

A wire of length 88 cm is bent to form a circle. When the wire is heated, the length increases at the rate of 0.3 cms^-1.

(a) Calculate the rate of change in the radius of the circle.

(b) Hence, calculate the radius of the circle after 5s.

Solution:

$\begin{array}{l} Length of circumference of a circle, \\ L = 2 π r \\ \frac{d L}{d r} = 2 π \end{array}$

(a)

$\begin{array}{l} Given \frac{d L}{d t} = 0.3 \\ Rate of change in the radius of the circle = \frac{d r}{d t} \\ \frac{d r}{d t} = \frac{d r}{d L} \times \frac{d L}{d t} \\ \frac{d r}{d t} = \frac{1}{2 π} \times 0.3 \\ \frac{d r}{d t} = 0.0477 {cms}^{- 1} \end{array}$

(b)

$\begin{array}{l} 2 π r = 88 \\ r = \frac{88}{2 π} = \frac{44}{π} \\ Hence, the radius of the circle after 5 s \\ = \frac{44}{π} + 5 (0.0477) \\ = 14.24 cm \end{array}$

Question 22:

The diagram shows a conical container with diameter 0.8m and height 0.6m. Water is poured into the container at a constant rate of 0.02m³s^-1. Calculate the rate of change of the height of the water level when the height of water level is 0.5m.

Solution:

$\begin{array}{l} Let, \\ h = height of the water level \\ r = radius of the water surface \\ V = volume of the water \\ \frac{r}{h} = \frac{0.4}{0.6} \leftarrow Concept of similar triangles \\ \frac{r}{h} = \frac{2}{3} \\ r = \frac{2}{3} h \\ Volume of water, V = \frac{1}{3} π r^{2} h \\ V = \frac{1}{3} π {(\frac{2}{3} h)}^{2} h \\ V = \frac{4}{27} π h^{3} \\ \frac{d V}{d h} = (3) \frac{4}{27} π h^{2} \\ \frac{d V}{d h} = \frac{4}{9} π h^{2} \\ The rate of change of the height of the water level \\ when the height of water level is 0 .5 m = \frac{d h}{d t} . \\ \frac{d h}{d t} = \frac{d h}{d V} \times \frac{d V}{d t} \leftarrow Chain rule \\ \frac{d h}{d t} = \frac{9}{4 π h^{2}} \times 0.02 \leftarrow Given \frac{d V}{d t} = 0.02 \\ \frac{d h}{d t} = \frac{9}{4 π {(0.5)}^{2}} \times 0.02 \\ \frac{d h}{d t} = 0.0572 m s^{- 1} \end{array}$