2 Marks

Question 12 Marks

Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2cm.

Answer

Let V be the volume of the sphere. Then,
$\text{V}=\frac{4}{3}\pi\text{r}^3$
$\Rightarrow\frac{\text{dv}}{\text{dr}}=4\pi\text{r}^2$
Let S be the total surface area of sphere. Then,
$\text{S}=4\pi\text{r}^2$
$\Rightarrow\frac{\text{dS}}{\text{dr}}=8\pi\text{r}$
$\therefore\frac{\text{dv}}{\text{dS}}=\frac{\text{dv}}{\text{dS}}/\frac{\text{dS}}{\text{dr}}$
$\Rightarrow\frac{\text{dV}}{\text{dS}}=\frac{4\pi\text{r}^2}{8\pi\text{r}}=\frac{\text{r}}{2}$
$\Rightarrow\Big(\frac{\text{dV}}{\text{dS}}\Big)_{\text{r}=2}=\frac{2}{2}$
$=1\text{cm}$

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Question 22 Marks

If a particle moves in a straight line such that the distance travelled in time t is given by s = t³ - 6t² + 9t + 8. Find the initial velocity of the particle.

Answer

$\text{S}=\text{t}^3-6\text{t}+9\text{t}+8$
$\frac{\text{dS}}{\text{dt}}=3\text{t}^2-12\text{t}+9$
Initial velocity t = 0
$\frac{\text{dS}}{\text{dt}}=3(0)^2-12(0)+9$
$\frac{\text{dS}}{\text{dt}}=9$ units/ units time.

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Question 32 Marks

Find the rate of change of the volume of a cone with respect to the radius of its base.

Answer

Let r be the radius, v be the volume of cone and h height.
$\text{V}=\frac{1}{3}\pi\text{r}^2\text{h}$
$\frac{\text{dv}}{\text{dr}}=\frac{1}{3}\pi\text{r}^2\text{h}.$

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Question 42 Marks

The sides of an equilateral triangle are increasing at the rate of 2cm/ sec. How far is the area increasing when the side is 10cms?

Answer

Let x be the side and A be the area of the equilateral triangle at any time t. Then,
$\text{A}=\frac{\sqrt{3}}{4}\text{x}^2$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=2\times\frac{\sqrt{3}}{4}\text{x}\frac{\text{dx}}{\text{dt}}$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=\frac{\sqrt{3}}{4}\times2\times10$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=10\sqrt{3}\text{cm}^2/\sec$

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Question 52 Marks

Find the surface area of a sphere when its volume is changing at the same rate as its radius.

Answer

Let r be the radius and V be the volume of the sphere at any time t then,

$\text{V}=\frac{4}{3}\pi\text{r}^3$

Implies that $\frac{\text{dV}}{\text{dt}}=4\pi\text{r}^2\Big(\frac{\text{dr}}{\text{dt}}\Big)$

Implies that $\frac{\text{dV}}{\text{dt}}=4\pi\text{r}^2\Big(\frac{\text{dV}}{\text{dt}}\Big)$ $\Big[\therefore\frac{\text{dV}}{\text{dt}}=\frac{\text{dr}}{\text{dt}}\Big]$

Implies that $4\pi\text{r}^2=1$

Implies that that surface area of sphere = 1 square unit.

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Question 62 Marks

Find an angle $\theta$
Whose rate of increase twice is twice the rate of decrease of its cosine.

Answer

$\frac{\text{d}\theta}{\text{dt}}=-2\frac{\text{d}}{\text{dt}}(\cos\theta)$
$\frac{\text{d}\theta}{\text{dt}}=-2(-\sin\theta)\frac{\text{d}\theta}{\text{dt}}$
$1=2\sin\theta$
$\sin\theta=\frac{1}{2}$
$\theta=\frac{\pi}{6}$

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Question 72 Marks

The side of a square is increasing at the rate of 0.1cm/ sec. Find the rate of increase of its perimeter.

Answer

Let x be the side and P be the perimeter of the square at any time t.
Then, $\text{P}=4\text{x}$
Implies that $\frac{\text{dP}}{\text{dt}}=4\frac{\text{dx}}{\text{dt}}$
Implies that $\frac{\text{dP}}{\text{dt}}=4\times0.1$
$\Big[\therefore\frac{\text{dx}}{\text{dt}}\text{dt}=0.1\text{cm}/\sec\Big]$
Implies that $\frac{\text{dP}}{\text{dt}}=0.4\text{cm}/\sec$

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Question 82 Marks

The total revenue received from the sale of x units of a product is given by R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.

Answer

Since the marginal revenue is the rate of change of total revenue with respect to its output,
Marginal Revenue (MR) $=\frac{\text{dR}(\text{x})}{\text{dx}}=\frac{\text{d}}{\text{dx}}(13\text{x}^2+26\text{x}+15)=26\text{x}+26$
When x = 7,
Marginal Revenue (MR) = 26(7) + 26 = 182 + 26 = Rs. 208

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Question 92 Marks

The side of an equilataral triangle is inreasing at the rate of $\frac{1}{3}\text{cm}/\sec.$ Find the rate of increase of his primeter.

Answer

Let x be the side and P be the perimeter of the equilateral triangle at any time t. Then,
$\text{P}=3\text{x}$
$\Rightarrow\frac{\text{dP}}{\text{dt}}=3\frac{\text{dx}}{\text{dt}}$
$\Rightarrow\frac{\text{dP}}{\text{dt}}=3\times\frac{1}{3}$$\Big[\therefore\frac{\text{dx}}{\text{dt}}=\frac{1}{3}\text{cm}/\sec\Big]$
$\Rightarrow\frac{\text{dP}}{\text{dt}}=1\text{cm}/\sec$

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Question 102 Marks

A stone is dropped into a quiet lake and waves move in circles at a speed of 4cm/ sec. At the instant when the radius of the circular wave is 10cm, how fast is the enclosed area increasing?

Answer

Let r be the radius and A be the area of the circle at any times t. Then,
$\text{A}=\pi\text{r}^2$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=2\pi\text{r}\frac{\text{dr}}{\text{dt}}$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=2\pi\times4\times10$ $\big[\therefore\text{r}=4\text{cm}\text{ and }\frac{\text{dr}}{\text{dt}}=10\text{cm}/\sec\big]$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=80\pi\text{ cm}^2/\sec$

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Question 112 Marks

The radius of a circle is increasing at the rate of 0.5cm/ sec. Find the rate of increase of its circumference.

Answer

Let r be radius and C be circumference of the circle.
Given, $\frac{\text{dr}}{\text{dt}}=0.5\text{cm}/\sec$
$\text{C}=2\pi\text{r}$
$\frac{\text{dC}}{\text{dt}}=2\pi\frac{\text{dr}}{\text{dt}}$
$=2\pi(0.5)$
$\frac{\text{dC}}{\text{dt}}=\pi\text{ cm}/\sec.$

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Question 122 Marks

Find the rate of change of the area of a circular disc with respect to its circumference when the radius is 3cm.

Answer

Let A be the area of the circular disc.
Then,
$\text{A}=\pi\text{r}^2$
Implies that $\frac{\text{dA}}{\text{dr}}=2\pi\text{r}$
Let C be the circumference of the circular disc.
Then,
$\text{C}=2\pi\text{r}$
Implies that $\frac{\text{dC}}{\text{dr}}=2\pi$
$\therefore\frac{\text{dA}}{\text{dC}}=\frac{\text{dA}/\text{dr}}{\text{dC}/\text{dr}}$
$\frac{\text{dA}}{\text{dC}}=\frac{2\pi\text{r}}{2\pi\text{r}}=\text{r}$
$\Big(\frac{\text{dA}}{\text{dC}}\Big)_\text{r=3}=3\text{cm}$

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Maths 2 Marks

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