3 Marks Question

Question 13 Marks

If $y = log_e x,$ then find $\triangle\text{y}$ when $x = 3$ and $\triangle\text{x} = 0.03.$

Answer

Here
$\text{x}=3,\triangle\text{x}=0.030\ \text{and}\ \text{y}=\log_\text{e}\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}-3}=\frac{1}{3}$
$\triangle\text{y}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}-3}\text{x}(\triangle\text{x})$
$=\Big(\frac{1}{3}\Big)(0.03)$
$\triangle\text{y}=0.01$

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Question 23 Marks

Find the approximate change in the value $V$ of a cube of side $x$ metres caused by increasing the side by $1\%.$

Answer

The volume of a cube $(V)$ of side $x$ is given by $V = x^3.$
$\therefore\text{DV}=\Big(\frac{\text{DV}}{\text{dx}}\Big)\triangle\text{x}$
$=(3\text{x}^2)\triangle\text{x}$
$=(3\text{x}^2)(0.01\text{x})\ [\text{as }1\%\text{ of x is }0.01\text{x}]$
$=0.03\text{x}^3$
Hence, the approximate change in the volume of the cube is $0.03x^3m^3.$

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Question 33 Marks

If the percentage error in the radius of a sphere is $\alpha,$ find the percentage error in its volume.

Answer

$\text{V}=\frac{4}{3}\pi\ \text{x}^3$
We have
$\frac{\triangle\text{x}}{\text{x}}\times100=\alpha$
$\Rightarrow\frac{\text{dV}}{\text{dx}}=4\pi\ \text{x}^2$
$\Rightarrow\frac{\text{dV}}{\text{V}}=\frac{4\pi\ \text{x}^2}{\text{V}}\text{dx}$
$\Rightarrow\frac{\triangle\text{V}}{\text{V}}=\frac{4\pi\ \text{x}^2}{\frac{4}{3}\pi\ \text{x}^3}\times\frac{\text{x}\alpha}{100}$
$\Rightarrow\frac{\triangle\text{V}}{\text{V}}\times100=3\alpha$
Hence, the percentage error in the volume is $3\alpha$

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Question 43 Marks

If $\text{y}=\sin\text{x}$ and x changes from $\frac{\pi}{2}$ to $\frac{22}{14}$, what is the approximate change in y?

Answer

Let:
$\text{x}=\frac{\pi}{2}$
$\text{x}+\triangle\text{x}=\frac{22}{14}$
$\Rightarrow\text{dx}=\triangle\text{x}=\frac{22}{14}=\frac{\pi}{2}=0$
Now, $\text{y}=\sin\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\cos\text{x}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{2}}=\cos\Big(\frac{\pi}{2}\Big)=0$
$\therefore\ \triangle\text{y}=\frac{\text{dy}}{\text{dx}}\Rightarrow\triangle\text{y}=0\text{x}=0\times0=0$
$\Rightarrow\triangle\text{y}=0$

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Question 53 Marks

A piece of ice is in the form of a cube melts so that the percentage error in the edge of cube is a, then find the percentage error in its volume.

Answer

Let, edge of the cube is x,
so,
$\triangle\text{x}=\text{a}\%\text{ of}\ \text{x}$
$\triangle\text{x}=(0.0\text{ ax})$
Now,
v = volume of sphere
$\text{v}=\text{x}^3$
$\frac{\text{dv}}{\text{dr}}=3\text{x}^2$
$\triangle\text{v}=\Big(\frac{\text{dv}}{\text{dr}}\Big)(\triangle\text{x})$
$=(3\text{x}^2)(0.0\text{ ax})$
$\triangle\text{v}=(3\text{x}^3)(0.0\ \text{a x})$
Percentage error in volume $=\frac{\triangle\text{v}\times100}{\text{v}}$
$=\frac{(3\text{x}^3)(0.0\ \text{a})\times100}{\text{x}^3}$
Percentage error in volume = 3a%

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