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Maths 4 Marks

Differentials, Errors and Approximations · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 41 questions.

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41 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
Using differentials, find the approximate values of the following:
$\Big(\frac{17}{81}\Big)^{\frac{1}{4}}$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=(\text{x})\frac{1}{4}$
Let:
$\text{x}=\frac{16}{81}$
$\text{x}+\triangle \text{x}=\frac{17}{81}$
Then,
$\triangle\text{x}=\frac{1}{81}$
For $\text{x}=\frac{16}{81},$
$\text{y}=\Big(\frac{16}{81}\Big)^{\frac{1}{4}}=\frac{2}{3}$
Let:
$\text{dx}=\triangle \text{x}=\frac{1}{81}$
Now, $\text{y}=(\text{x})^ {\frac{1}{4}}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{4(\text{x})^{\frac{3}{4}}}$
$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{16}{81}}=\frac{27}{32}$
$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {12}\times\frac{1}{81}=\frac{1}{96}=0.01042$
$\Rightarrow\triangle \text{y} =0.01042$
$\therefore\Big(\frac{17}{81}\Big)^{\frac{1}{4}}=\text{y}+\triangle\text{y} =0.6771$
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Question 24 Marks
Using differentials, find the approximate values of the following:
$(29)^{\frac{1}{3}}$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=(\text{x})^{\frac{1}{3}}$

Let:

$\text{x} =27$

$\text{x}+\triangle \text{x}=29$

Then,

$\text{x}= 2$

For $\text{x}=27$

$\text{y}=(27)^{\frac {1}{3}}=3$

Let:

$\text{dx}=\triangle \text{x}=2$

Now  $\text{y}=(\text {x})^{\frac{1}{3}}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{3(\text {x})^{\frac{2}{3}}}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}= 27}=\frac{1}{27}$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {27}\times2=0.074$

$\Rightarrow\triangle \text{y} =0.074$

$\therefore(29)^{\frac {1}{3}}=\text{y}+\triangle\text{y} =3.074$

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Question 34 Marks
Using differentials, find the approximate values of the following:
$(0.009)^{\frac{1}{3}}$
Answer
consider the function $=\text{f}(\text{x})=\sqrt[3]{\text{x}}$
Let:
x = 0.008
$\text{x}+\triangle\text{x}=0.009$
then, $\triangle\text{x}=0.001$
For x = 0.008,
$\text{y}=\sqrt{0.008}=0.2$
Let:
$\text{d}+\triangle\text{d}=0.001$
Now $=\text{y}=\sqrt[3]{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{3(\text{x})^{\frac{2}{3}}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{d}}\Big)_{\text{x}=0.008}=\frac{1}{3\times0.04}-\frac{1}{0.12}$
$\therefore\triangle\text{y}=\frac{1}{120}=0.008333$
$\therefore(0.009)^{\frac{1}{3}}=\text{y}+\triangle\text{y}=0.208333$
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Question 44 Marks
Find the approximate value of log10 1005, given that log10 e = 0.4343.
Answer
Let:
$\text{y}=\text{f}(\text{x})=\log_{10}\text{x}$
here,
x = 1000,
$\text{x}+\triangle\text{x}=1005$
$\Rightarrow\triangle\text{x}=5$
$\Rightarrow\text{dx}=\triangle\text{x}=5$
For x = 1000,
$\text{y}=\log_{10}1000=\log_{10}(10)^3=3$
Now, $\text{y}=\log_{10}\text{x}=\frac{\log_\text{e}\text{x}}{\log_\text{e}10}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{0.4343}{\text{x}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=1000}=\frac{0.4343}{1000}=0.0004343$
$\triangle\text{y}=\text{dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=0.0004343\times5=0.0021715$
$\therefore\log_{10}1005=\text{y}+\triangle\text{y}=3.0021715$
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Question 54 Marks
If the radius of a sphere is measured as 9cm with an error of 0.03m, find the approximate error in calculating its surface area.
Answer
Let r be the radius of the sphere and $\triangle\text{r}$ be the erroe in measuring the radius.
then,
r = 9m and $\triangle\text{r}=0.03\text{m}$
Now, the surface area of the sphere (S) is given by,
$\text{S}=4\pi\text{r}^2$
$\therefore\frac{\text{DS}}{\text{dr}}=8\pi\text{r}$
$\therefore\text{DS}=\Big(\frac{\text{dS}}{\text{dr}}\Big)\triangle\text{r}$
$=(8\pi\text{r})\triangle\text{r}$
$=8\pi\text{r}(9)(0.03)\text{m}^3$
$=2.16\pi\ \text{m}^2$
Hence, the approximate error in calculating the surface area is $2.16\pi\ \text{m}^2$
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Question 64 Marks
Using differentials, find the approximate values of the following:
$\sqrt{26}$
Answer
Let $\text{x}=25,\\\text{x}+ \triangle\text{x}=26$

$\triangle\text{x}=26- 25$

$=-1$

Let $\text{y}=\sqrt{\text {x}}$

$\frac{\text{dy}} {\text{dx}}=\frac{1}{2\sqrt{\text{x}}}$

$\Big(\frac{\text{dy}} {\text{dx}}\Big)_{\text{x}=25}=\frac{1} {2\sqrt{25}}$

$=\frac{1}{10}$

$=0.1$

$\triangle\text{y}= \Big(\frac{\text{dy}}{\text{dx}}\Big)_ {\text{x}=25}\times(\triangle\text{x})$

$=(0.01)(1)$

$=0.01$

$\sqrt{26}=\text{y}+ \triangle\text{y}$

$=\sqrt{\text {x}}+0.01$

$\sqrt{25}+0.1$

$\sqrt{26}=5.1$

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Question 74 Marks
Using differentials, find the approximate values of the following:
$(255)^{\frac{1}{4}}$
Answer
Let $\text{x}=256,\text {x}+\triangle\text{x}=255$

$\triangle\text{x}=255 -256$

$\triangle\text{x}=1$

Let $\text{y}=\text{x}^ {\frac{1}{4}}$

$\frac{\text{dy}} {\text{dx}}=\frac{1}{4\text{x}^{\frac{3} {4}}}$

$\Big(\frac{\text{dy}} {\text{dx}}\Big)_{\text{x}\Rightarrow256}=\frac{1} {4(256)^{\frac{3}{4}}}$

$=\frac{1}{256}$

$=0.00391$

Now,

$\triangle\text{y}= \Big(\frac{\text{dy}}{\text{dx}}\Big)_ {\text{x}-256}\times\triangle\text{x}$

$=(0.00391)(-1)$

$\triangle\text{y}=- 0.00391$

$(255)^{\frac{1}{4}}= \text{y}+\triangle\text{y}$

$=(\text{x})^{\frac{1} {4}}+(-0.00391)$

$=4-0.00391$

$(255)^{\frac{1}{4}} =3.99609$

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Question 84 Marks
Using differentials, find the approximate values of the following:
$(82)^{\frac{1}{4}}$
Answer
Let $\text{x}=81,\\\text{x}+ \triangle\text{x}=82$

$\triangle\text{x}=82- 81$

$=1$

Let $\text{y}=\text{x}^ {\frac{1}{4}}$

$\frac{\text{dy}} {\text{dx}}=\frac{1}{4\text{x}^{\frac{3} {4}}}$

$\Big(\frac{\text{dy}} {\text{dx}}\Big)_{\text{x}=81}=\frac{1} {4(81)^{\frac{3}{4}}}$

$=\frac{1}{108}$

$=0.009259$

$\triangle\text{y}= \Big(\frac{\text{dy}}{\text{dx}}\Big)_ {\text{x}=81}\times(\triangle\text{x})$

$=(0.0008259)(1)$

$=0.009259$

$(82)^{\frac{1}{4}}= \text{y}+ \triangle\text{y}$

$=\text {x}^\frac{1} {4}+0.009259$

$=(81)\frac{1} {4}+0.009259$

$=(81)\frac{1} {4}+3.009259$

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Question 94 Marks
Using differentials, find the approximate values of the following:
$(80)^{\frac{1}{4}}$
Answer
Let $\text{x}=81,\\\text{ x}+ \triangle\text{x}=80$

$\triangle\text{x}=80- 81$

$=-1$

Let $\text{y}=\text{x}^ {\frac{1}{4}}$

$\frac{\text{dy}} {\text{dx}}=\frac{1}{4(81)^{\frac{3} {4}}}$

$=\frac{1}{108}$

$0.00926$

$\triangle\text{y}= \Big(\frac{\text{dy}}{\text{dx}}\Big)_ {\text{x}=81}\times(\triangle\text{x})$

$=0.00926)(-1)$

$=-0.00926$

$(80)^{\frac{1}{4}}= \text{y}+\triangle\text{y}$

$=\text{x}^{\frac{1} {4}}-0.00926$

$=(81)^{\frac{1}{4}}- 0.00926$

$3-0.0026$

$2.9974$

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Question 104 Marks
Using differentials, find the approximate values of the following:

$\frac{1}{\sqrt{25.1}}$

Answer
Let $\text{x}=25,\text{x}+ \triangle\text{x}=251.$

$\triangle\text{x} =25.1-25$

$\triangle\text{x} =0.1$

Let $\text{y}=\frac{1} {\sqrt{\text{x}}}$

$\frac{\text{dy}} {\text{dx}}=\frac{2}{\text{x}^{\frac{3} {2}}}$

$\Big(\frac{\text{dy}} {\text{dx}}\Big)_{\text{x}=25}=-\frac{1} {2(25)^{\frac{3}{2}}}$

$=-\frac{1}{250}$

$=-0.004$

Now,

$\triangle\text{y}= \Big(\frac{\text{dy}}{\text{dx}}\Big)_ {\text{x}=25}\times(\triangle\text{x})$

$=(-0.004)(0.1)$

$=-0.0004$

$\frac{1}{\sqrt{25.1}} =\text{y}+\triangle\text{y}$

$=\frac{1}{\sqrt{\text {x}}}+(-0.0004)$

$=\frac{1}{\sqrt{25}}- 0.0004$

$=\frac{1}{5}-0.0004$

$=0.2-0.0004$

$\frac{1}{\sqrt{25.1}} =0.1996$

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Question 114 Marks
Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2
Answer
Let x = 2 and $\triangle\text{x}=0.01$ then we have:

$\text{f}(2.01)=\text{f}(\text{x}+\triangle\text{x})=4(\text{x}+\triangle\text{x}^2)+5(\text{x}+\triangle\text{x})+2$

Now, $\triangle\text{y}=\text{f}(\text{x}+\triangle\text{x})-\text{f}(\text{x})$

$\therefore\text{f}(\text{x}+\triangle\text{x})=\text{f}(\text{x})+\triangle\text{y}$

$\approx\text{f}(\text{x})+\text{f}'(\text{x}).\triangle\text{x}$

$\Rightarrow\text{f}(2.01)\approx(4\text{x}^2+5\text{x}+2)+(8\text{x}+5)\triangle\text{x}$

$=4(2)^2+5(2)+2]+[8(2)+5](0.01)$

$=(16+10+2)+(16+5)(0.01)$

$=28+(21)(0.01)$

$=28+0.21$

$=28.21$

Hence, the approximate value of f (2.01) is 28.21

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Question 124 Marks
Find the approximate value of (5.001), where f(x) = x3 − 7x2 + 15.
Answer
Let:
x = 5
$\text{x}+\triangle\text{x}=5.001$
$\Rightarrow\triangle\text{x}=0.001$
$\text{f}(\text{x})=\text{x}^3-7\text{x}^2+15$
$\Rightarrow\text{y}=\text{f}(\text{x}=3)=125-175+15=-35$
Now, y = f (x)
$\Rightarrow\frac{\text{dy}}{\text{dx}}=3\text{x}^2-14\text{x}$
$\therefore\text{dy}=\triangle\text{y}=\frac{\text{dy}}{\text{dx}}\text{dx}=(3\text{x}^2-14\text{x})\times0.001=(75-70)\times0.001=0.005$
$\therefore\text{f}(5.001)=\text{y}+\triangle\text{y}=-35+0.005=-34.995$
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Question 134 Marks
Using differentials, find the approximate values of the following:
$\sqrt{49.5}$
Answer
Define a function $\text{y}=\text{x}^{\frac{3}{2}}$
For x = 4, y = 8
$\text{x}+\triangle\text{x}=3.968\\\Rightarrow\triangle\text{x}=3.968-4=-0.032$
$\frac{\text{dy}}{\text{dx}}=\frac{3}{2}\text{x}^{\frac{1}{2}}$
$\text{dy}=\Big(\frac{3}{2}\text{x}^{\frac{1}{2}}\Big)\text{dx}$
$\Rightarrow\triangle\text{y}\mid_{\text{x}=4}\simeq(3)\triangle\text{x}$
$\Rightarrow\triangle\text{y}\mid_{\text{x}=4}\simeq(3)\times(-0.032)=-0.096$
$(3.968)^{\frac{3}{2}}=\text{y}+\triangle\text{y}=8-0.096$
$=7.904$
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Question 144 Marks
Using differentials, find the approximate values of the following:
$\sqrt{49.5}$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=\sqrt{\text{x}}$
Let:
$\text{x} =49$
$\text{x}+\triangle \text{x}=49.5$
Then,
$\triangle\text{x}=0.5$
For $\text{x}=49$
$\text{y}=\sqrt{49}=7$
Let:
$\text{dx}=\triangle \text{x}=0.5$
Now, $\text{y}=(\text{x})^ {\frac{1}{2}}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}$
$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =49}=\frac{1}{14}$
$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {14}\times0.5=0.0357$
$\Rightarrow\triangle \text{y} =0.0357$
$\therefore\sqrt{49.5} =\text{y}+\triangle\text{y} =7.0357$
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Question 154 Marks
A circular metal plate expends under heating so that its radius increases by k%. Find the approximate increase in the area of the plate, if the radius of the plate before heating is 10cm.
Answer
Let at any time, x be the radius and y be the area of the plate.
$\text{y}=\text{x}^2$
Let $\triangle\text{x}$ be the change in the radius and $\triangle\text{y}$ be the change in the area of the plate.
We have,
$\frac{\triangle\text{x}}{\text{x}}\times100=\text{k}$
when x = 10, we get
$\triangle\text{x}=\frac{10\text{k}}{100}=\frac{\text{k}}{10}$
Now, $\text{y}=\pi\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\pi\text{x}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=10\text{cm}}=20\pi\ \text{cm}^2/\text{cm}$
$\therefore\triangle\text{y}=\text{dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=20\pi\times\frac{\text{k}}{10}=2\text{k}\pi\ \text{cm}^2$
Hence, the approximate change in the area of the plate is $2\text{k}\pi\ \text{cm}^2$
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Question 164 Marks
Using differentials, find the approximate values of the following:
$\sqrt{0.082}$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=\sqrt{\text{x}}$

Let:

$\text{x}=0.0841$

$\text{x}+\triangle \text{x}=0.082$

Then,

$\triangle\text{x}=0.0021$

For $\text{x}=0.0841$

$\text{y}=\sqrt{0.0841}=0.29$

Let:

$\text{dx}=\triangle \text{x}=-0.0021$

Now, $\text{y}=(\text{x})^{\frac{1}{2}}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}=0.0841}=\frac{1}{0.58}=\frac{50}{29}$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{50}{29}\times(-0.0021)=-0.0036$

$\Rightarrow\triangle\text{y} =-0.0036$

$\therefore\sqrt{0.082}=\text{y}+\triangle\text{y}=0.2864$

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Question 174 Marks
Using differentials, find the approximate values of the following:
$(15)^{\frac{1}{4}}$
Answer
consider the function $\text{y}=\text{f}(\text{x})=\text{x}^{\frac{1}{4}}$
Let:
x = 16
$\text{x}+\triangle\text{x}=15$
Then,
$\triangle\text{x}=-1$
For x = 16
$\text{y}=(16)^{\frac{1}{4}}=2$
Let:
$\text{dx}=\triangle\text{x}=-1$
Now, $\text{y}=(\text{x})^{\frac{1}{4}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{4(\text{x})^{\frac{3}{4}}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=16}=\frac{1}{32}$
$\therefore\triangle\text{y}=\text{dy}=\Big(\frac{\text{dy}}{\text{dx}}\Big)\text{dx}=\frac{1}{32}\times(-1)=\frac{-1}{32}$
$\Rightarrow\triangle\text{y}=\frac{-1}{32}=-0.03125$
$\therefore(15)^{\frac{1}{4}}=\text{y}+\triangle\text{y}=1.96875$
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Question 184 Marks
If the radius of a sphere is measured as 7m with an error of 0.02m, find the approximate error in calculating its volume.
Answer
$\text{y}=\frac{4}{3}\pi\text{x}^3$
Let $\triangle\text{x}$ be the error in the radius
x = 7
$\triangle\text{x}=0.02$
$\frac{\text{dy}}{\text{dx}}=4\pi\text{x}^2$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=7}=196\pi$
$\therefore\triangle\text{y}=\text{dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=196\pi\times0.02=3.92\pi$
Hence, the approximate error in calculating the volume of the sphere is $3.92\pi\ \text{m}^3$
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Question 194 Marks
Using differentials, find the approximate values of the following:
$\sqrt{401}$
Answer
Consider the function $\text{y}=\text{f}(\text{x})=\sqrt{\text{x}}$
Let:
x = 400
$\text{x}+\triangle\text{x}=401$
Then,
$\triangle\text{x}=1$
For x = 400
$\text{y}=\sqrt{400}=20$
Let:
$\text{dx}=\triangle\text{x}=1$
Now, $\text{y}=\sqrt{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=400}=\frac{1}{40}$
$\therefore\triangle\text{y}=\text{dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=\frac{1}{40}\times1=\frac{1}{40}$
$\Rightarrow\triangle\text{y}=\frac{1}{40}=0.025$
$\therefore\sqrt{401}=\text{y}=\triangle\text{y}=20.025$
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Question 204 Marks
Using differentials, find the approximate values of the following:
$\cos61^\circ$ it being given that $\sin60^\circ=0.86603$ and $0.01745$ radian
Answer
Consider the function $\text{y}=\text{f} (\text{x})=\cos\text{x}^\circ$

Let:

x = 60°

$\text{x}+\triangle \text{x}=61^\circ$

Then,

$\triangle\text{x}=1^ \circ=0.01745$

For $\text{x}=60^\circ$

$\text{y}\cos60^\circ- 0.5$

Let:

$\text{dx}=\triangle \text{x}=0.01745$

Now, $\text{y}=\cos\text {x}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=-\sin\text{x}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =60}=-0.86603$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=- 0.86603\times0.01745=-0.01511$

$\Rightarrow\text{y} =0.01511$

$\therefore\cos61^ \circ=\text{y}+\triangle\text{y} =0.48488\approx0.48489$

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Question 214 Marks
The pressure p and the volume v of a gas are connected by the relation pv1.4 = const. Find the percentage error in p corresponding to a decrease of 1/2% in v.
Answer
We have
$\text{pv}^{1.4}=\text{constant}=\text{k}(\text{say})$
Taking log on both the sides, we get
$\log(\text{pv}^{1.4})=\log\text{k}$
$\Rightarrow\log\text{p}+1.4\log\text{v}=\log\text{k}$
Differentiating both the sides w.r.t. x we get
$\frac{1}{\text{p}}\frac{\text{dp}}{\text{dv}}+\frac{1.4}{\text{v}}=0$
$\Rightarrow\frac{\text{dp}}{\text{p}}=\frac{-1.4\text{dv}}{\text{v}}$
Now, $\text{dp}=\frac{\text{dp}}{\text{dv}}\text{dv}=\frac{1.4\text{p}}{\text{v}}\text{dv}$
$\Rightarrow\frac{\text{dp}}{\text{p}}\times100=-1.4\Big(\frac{\text{dp}}{\text{v}}\times100\Big)\\=-1.4\times\Big(\frac{-1}{2}\Big)=0.7$ [since we are given $\frac{1}{2}\%$ decrease in v]
Hence the error in p is 0.7%
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Question 224 Marks
The height of a cone increases by k%, its semi-vertical angle remaining the same. What is the approximate percentage increase
  1. In total surface area, and
  2. In the volume, assuming that k is small?
Answer
Let $\triangle\text{h}$ be the change in the height, $\triangle\text{r}$ be the change in the radius of base and $\triangle\text{l}$ be the change in the slant height.

semi-vertical angle remaining the same.

$\therefore\frac{\triangle\text{h}}{\text{h}}=\frac{\triangle\text{r}}{\text{r}}=\frac{\triangle\text{l}}{\text{l}}$

Also, $\frac{\triangle\text{h}}{\text{h}}\times100=\text{k}$

Then, $\frac{\triangle\text{h}}{\text{h}}\times100=\frac{\triangle\text{r}}{\text{r}}\times100=\frac{\triangle\text{l}}{\text{l}}\times100=\text{k}...(1)$

  1. Total surface area of the cone, $\text{T}=\pi\text{rl}+\pi\text{r}^2$

Differentiating both sides w.r.t.

we get

$\frac{\text{dt}}{\text{dr}}=\pi\text{l}+\pi\text{r}\frac{\text{dl}}{\text{dr}}+2\pi\text{r}$

$\frac{\text{dt}}{\text{dr}}=\pi\text{l}+\pi\text{r}\frac{\text{l}}{\text{r}}+2\pi\text{r}\Big[\text{from}(1),\frac{\text{dl}}{\text{dr}}=\frac{\triangle\text{l}}{\triangle\text{r}}=\frac{\text{l}}{\text{r}}\Big]$

$\Rightarrow\frac{\text{dt}}{\text{dr}}=\pi\text{l}+\pi\text{l}+2\pi\text{r}$

$\Rightarrow\frac{\text{dt}}{\text{dr}}=2\pi(\text{l}+\text{r})$

$\therefore\triangle\text{T}=\frac{\text{dt}}{\text{dr}}\triangle\text{r}=2\pi(\text{l}+\text{r})\times\frac{\text{kr}}{100}=\frac{2\text{kr}\pi(\text{l}+\text{r})}{100}$

$\frac{\triangle\text{T}}{\text{T}}\times100=\frac{\Big(\frac{2\text{kr}\pi(\text{l}+\text{r})}{100}\Big)}{2\pi(\text{l}+\text{r})}\times100=2\text{k}\%$

Hence, the percentage increase in total surface area

$\frac{\text{dT}}{\text{dr}}=\pi\text{l}+\pi\text{r}\frac{\text{dl}}{\text{dr}}+2\pi\text{r}$

$\Rightarrow\frac{\text{dT}}{\text{dr}}=\pi\text{l}+\pi\text{r}\frac{\text{l}}{\text{r}}+2\pi\text{r}$

$\Rightarrow\frac{\text{dT}}{\text{dr}}=2\pi(\text{l}+\text{r})$

$\triangle\text{T}=\frac{\text{dT}}{\text{dr}}\triangle\text{r}=2\pi(\text{l}+\text{r})\times\frac{\text{kr}}{100}=\frac{2\text{kr}\pi(\text{l}+\text{r})}{100}$

$\frac{\triangle\text{T}}{\text{T}}\times100=\frac{\frac{2\text{kr}\pi(\text{l}+\text{r})}{100}}{2\pi\text{r}(\text{l}+\text{r})}\times100=2\text{k}\%$

  1. Let v = volume of cone

$=\frac{1}{3}\pi\text{r}^2\text{h}$

$=\frac{1}{3}\pi(\text{h}\tan\alpha)^2\text{h}$

$\text{v}=\frac{\pi}{3}\tan^2\alpha\text{h}^2$

Differentiating it with respect to h treating $\alpha$ as constant,

$\frac{\text{dv}}{\text{dh}}=\pi\tan^2\alpha\times\text{h}$

$\triangle\text{v}=\Big(\frac{\text{dv}}{\text{dh}}\Big)\triangle\text{h}$

$=\pi\tan^2\alpha\text{h}^2\times(0.0\text{kh})$

$\triangle\text{v}=0.0\text{k}\pi\text{h}^3\tan^2\alpha$

Percentage increase in v $=\frac{\triangle\text{v}\times100}{\text{v}}$

$=\frac{0.0\text{k}\pi\text{h}^3\tan^2\alpha\times100}{\frac{\pi}{3}\tan^2\alpha\text{h}^3}$

$=3\text{k}\%$

So, percentage increase in volume $=3\text{k}\%$

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Question 234 Marks
Using differentials, find the approximate values of the following:
$\log_\text{e}4.04,$ it being given that $\log_{{10}^{4}}=0.6021$ and $\log_{10}\text{e}=0.4343$
Answer
Consider the function  $\text{y}=\text{f}(\text{x})\log_\text {e}\text{x}.$

Let:

x = 4

$\text{x}+\triangle \text{x}=4.04$

Then,

$\triangle\text{x} =0.04$

For x = 4

$\text{y}=\log_\text {e}4=\frac{\log_{10}4}{\log_{10}\text {e}}=\frac{0.6021}{0.4343}=1.386368$

Let:

$\text{dx}=\triangle \text{x}=0.04$

Now, $\text{y}=\log_\text {e}\text{x}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{\text {x}}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}=4} =\frac{1}{4}$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {4}\times0.04=0.01$

$\Rightarrow\triangle \text{y}=0.01$

$\therefore\log_\text {e}4.04=\text{y}+\triangle\text{y} =1.39638$

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Question 244 Marks
Using differentials, find the approximate values of the following:
$\frac{1}{(2.002)^2}$
Answer
consider the function $\text{y}=\text{f} (\text{x})=\frac{1}{\text{x}^2}$

Let:

x = 2

$\text{x}+\triangle \text{x}=2.002$

Then,

$\triangle\text{x}=- 0.002$

For x = 2

$\text{y}=\frac{1} {2^2}=\frac{1}{4}$

Let:

$\text{dx}=\triangle \text{x}=0.002$

Now, $\text{y}=\frac{1} {\text{x}^2}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{2}{\text{x} ^3}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}=2} =\frac{1}{4}$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1}{4}\times- 0.002=-0.005$

$\Rightarrow\triangle \text{y}=-0.0005$

$\therefore\frac{1} {(2.002)^2}=\text{y}+\triangle\text{y} =0.2495$

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Question 254 Marks
Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.
Answer
Let y be the surface area of the cube.
$\text{y}=6\text{x}^2$
We have
$\frac{\triangle\text{x}}{\text{x}}\times100=1$
Now,
$\frac{\text{dy}}{\text{dx}}=12\text{x}$
$\Rightarrow\triangle\text{y}=\text{dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=12\text{x}\times\frac{\text{x}}{100}=0.12\text{x}^2\text{m}^2$
hence, approximate change in the surface area of the cube is 0.14x2m2
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Question 264 Marks
Using differentials, find the approximate values of the following:
$(33)^{\frac{1}{5}}$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=(\text{x})^{\frac{1}{5}}$
Let:
$\text{x} =32$
$\text{x}+\triangle \text{x}=33$
Then,
$\triangle\text{x}=1$
For $\text{x}=33$
$\text{y}=(32)^{\frac{1}{5}}=2$
Let:
$\text{dx}=\triangle \text{x}=1$
Now, $\text{y}=(\text{x})^ {\frac{1}{5}}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{5(\text{x})^{\frac{4}{5}}}$
$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =32}=\frac{1}{80}$
$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {80}\times(1)=0.0125$
$\Rightarrow\triangle \text{y} =0.0125$
$\therefore(33)^{\frac{1}{5}} =\text{y}+\triangle\text{y} =2.0125$
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Question 274 Marks
Find the percentage error in calculating the surface area of a cubical box if an error of 1% is made in measuring the lengths of edges of the cube.
Answer
Let x be the edge of the cube and y be the surface area.
y = x2
Let $\triangle\text{x}$ be the error in x and $\triangle\text{y}$ be the corresponding error in y.
We have
$\frac{\triangle\text{x}}{\text{x}}\times100=1$
$\Rightarrow2\text{x}=\frac{\text{x}}{100}$
Now, y = x2
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\text{x}$
$\therefore\triangle\text{y}=\frac{\text{dy}}{\text{dx}}\times\triangle\text{x}=2\text{x}\times\frac{\text{x}}{100}$
$\Rightarrow\triangle\text{y}=2\frac{\text{x}^2}{100}$
$\Rightarrow\triangle\text{y}=2\frac{\text{y}}{100}$
$\Rightarrow\frac{\triangle\text{y}}{\text{y}}=\frac{2}{100}$
$\Rightarrow\frac{\triangle\text{y}}{\text{y}}\times100=2$
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Question 284 Marks
Using differentials, find the approximate values of the following:
$\log_\text{e}10.02$ it being given that $\log_\text{e}10=2.3026$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=\log_\text{e}\text{x}$

Let:

x = 10

$\text{x}+\triangle \text{x}=10.02$

Then,

$\triangle\text{x} =0.02$

For x

$\text{y}\log_\text{e} 10=2.3026$

Let:

$\text{dx}=\triangle \text{x}=0.02$

Now, $\text{y}=\log_\text {e}\text{x}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{\text {x}}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =10}=\frac{1}{10}$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {10}\times0.02=0.002$

$\Rightarrow\text{y} =0.002$

$\therefore\log_\text {e}10.02=\text{y}+\triangle\text{y} =2.3046$

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Question 294 Marks
Using differentials, find the approximate values of the following:
$\sqrt{37}$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=\sqrt{\text{x}}$
Let:
$\text{x} =36$
$\text{x}+\triangle \text{x}=37$
Then,
$\triangle\text{x}=1$
For $\text{x}=36$
$\text{y}=\sqrt{36} =6$
Let:
$\text{dx}=\triangle \text{x}=1$
Now  $\text{y}=(\text{x})^ {\frac{1}{2}}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{2\sqrt {\text{x}}}$
$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =36}=\frac{1}{12}$
$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {12}\times1=0.0833$
$\Rightarrow\triangle \text{y} =0.0833$
$\therefore\sqrt{37}= \text{y}+\triangle\text{y} =6.0833$
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Question 304 Marks
Show that the relative error in computing the volume of a sphere, due to an error in measuring the radius, is approximately equal to three times the relative error in the radius.
Answer
Let error in radius (r) = x% of r
$\triangle\text{r}=0.0\text{xr}$
Let v = volume of sphere
$\text{v}=\frac{4}{3}\pi\text{r}^3$
Differentiating it with respect to r,
$\frac{\text{dv}}{\text{dr}}=4\pi\text{r}^2$
so,
$\triangle\text{v}=\Big(\frac{\text{dv}}{\text{dr}}\Big)\times\triangle\text{r}$
$=(4\pi\text{r}^2)(0.0\text{x})\text{r}$
$\triangle\text{v}=0.0\text{x}\times4\pi\text{r}^3$
percentage of error in volume $\frac{\triangle\text{v}\times100}{\text{v}}$
$=\frac{(0.0\text{x})4\pi\text{r}^3\times100}{\frac{4}{3}\pi\text{r}^3}$
$=3\text{x}\%$
percentage of error in volume = 3(percentage of error in radius).
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Question 314 Marks
The radius of a sphere shrinks from 10 to 9.8cm. Find approximately the decrease in its volume.
Answer
Let $\text{x}=10,\\\text{x}+\triangle\text{x}=9.8$
$\triangle\text{x}=9.8-\text{x}$
$=9.8-10$
$\triangle\text{x}=-0.2$
$\text{y}=\frac{4}{3}\pi\text{x}^3$
$\frac{\text{dy}}{\text{dx}}=4\pi\text{r}^2$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}\Rightarrow10}=4\pi(10)^2$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}\Rightarrow10}=400\pi\ \text{cm}^2$
$=400\pi\times(-0.2)$
$\triangle\text{y}=-80\pi\ \text{cm}^3$
so, approximate decrease in volume is $80\pi\ \text{cm}^3$
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Question 324 Marks
Using differentials, find the approximate values of the following:
$\sin\Big(\frac{22}{14}\Big)$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=\sin\text{x}^\circ$

Let:

$\text{x}=\frac{22}{7}$

$\text{x}+\triangle \text{x}=\frac{22}{14}$

Then,

$\triangle\text{x}= \frac{-22}{14}$

For $\text{x}=\pi$

$\text{y}=\sin\Big (\frac{22}{7}\Big)=0$

Let:

$\text{dx}=\triangle \text{x}=\sin\frac{-22}{14}=-\sin\Big (\frac{\pi}{2}\Big)=-1$

Now, $\text{y}=\sin\text {x}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=\cos\text{x}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}= \frac{22}{7}}=-1$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=-1\times(-1)=1$

$\Rightarrow\triangle \text{y} =1$

$\therefore\sin\frac {22}{14}=\text{y}+\triangle\text{y}=1$

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Question 334 Marks
Using differentials, find the approximate values of the following:
$\cos\Big(\frac{11\pi}{36}\Big)$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=\cos\text{x}$

Let:

$\text{x}=\frac{\pi}{3}$

$\text{x}+\triangle \text{x}=\frac{11\pi}{36}$

Then,

$\triangle\text{x}= \frac{-\pi}{36}=-5^\circ$

For $\text{x}=\frac{\pi} {3}$

$\text{y}=\cos\Big (\frac{\pi}{3}\Big)=0.5$

Let:

$\text{dx}=\triangle \text{x}=-\sin5^\circ=-0.08716$

Now $\text{y}=\cos\text {x}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=-\sin\text{x}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}= \frac{\pi}{3}}=0.86603$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=-0.86603\times(- 0.08716)=0.075575$

$\Rightarrow\triangle \text{y} =0.075575$

$\therefore\cos\frac {11\pi}{36}=\text{y}+\triangle\text{y} =0.5+0.075570=0.575575$

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Question 344 Marks
Using differentials, find the approximate values of the following:
$(66)^{\frac{1}{3}}$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=(\text{x})^{\frac{1}{3}}$

Let:

$\text{x}=64$

$\text{x}+\triangle \text{x}=66$

Then,

$\triangle\text{x}=2$

For $\text{x}=64$

$\text{y}=(64)^{\frac {1}{3}}=4$

Let:

$\text{dx}=\triangle \text{x}=2$

Now, $\text{y}=(\text{x})^ {\frac{1}{3}}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{3(\text {x})^{\frac{2}{3}}}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}} =64=\frac{1}{48}$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {48}\times2=0.042$

$\Rightarrow\triangle \text{y} =0.042$

$\therefore(66)^{\frac {1}{3}}=\text{y}+\triangle\text{y} =4.042$

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Question 354 Marks
Using differentials, find the approximate values of the following:
$\sqrt{25.02}$
Answer
consider the function $\text{y}=\text{f}(\text{x})=\sqrt{\text{x}}$
Let:
x = 25
$\text{x}+\triangle\text{x}=25.02$
Then
$\triangle\text{x}=0.02$
For x = 25
$\text{y}=\sqrt{25}=5$
Let:
$\text{dx}+\triangle\text{x}=0.02$
Now, $\text{y}=\sqrt{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=25}=\frac{1}{10}$
$\therefore\triangle\text{y}=\text{dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=\frac{1}{10}\times0.02=0.002$
$\Rightarrow\triangle\text{y}=0.002$
$\therefore\text{y}+\triangle\text{y}=5.002$
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Question 364 Marks
Using differentials, find the approximate values of the following:
$\log_\text{e}10.01$ it being given that $\log_{10}=0.4343$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=\log_{10}\text{x}$

Let:

x = 10

$\text{x}+\triangle \text{x}=10.01$

Then,

$\triangle\text{x} =0.01$

For x

$\text{y}\log_{10} 10=1$

Let:

$\text{dx}=\triangle \text{x}=0.01$

Now, $\text{y}=\log_ {10}\text{x}=\frac{\log_\text{e}\text {x}}{\log_\text{e}10}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1} {2.3025\text{x}}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =10}=0.04343$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx} =0.04343\times0.01=0.004343$

$\Rightarrow\text{y} =0.004343$

$\therefore\log_{10} 10.01=\text{y}+\triangle\text{y} =1.004343$

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Question 374 Marks
Using differentials, find the approximate values of the following:
$\sqrt{0.48}$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=\sqrt{\text{x}}$
Let:
$\text{x} =0.49$
$\text{x}+\triangle \text{x}=0.48$
Then,
$\triangle\text{x}=- 0.01$
For $\text{x}=0.49$
$\text{y}=\sqrt{0.49} =0.7$
Let:
$\text{dx}=\triangle \text{x}=0.01$
Now  $\text{y}=(\text{x})^ {\frac{1}{2}}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{2\sqrt {\text{x}}}$
$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =0.49}=\frac{1}{1.4}$
$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {1.4}\times(-0.01)=0.007143$
$\Rightarrow\triangle \text{y} =0.007143$
$\therefore\sqrt{0.48} =\text{y}+\triangle\text{y} =0.693$
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Question 384 Marks
Using differentials, find the approximate values of the following:
$25^{\frac{1}{3}}$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=(\text{x})^{\frac{1}{3}}$

Let:

$\text{x}=27$

$\text{x}+\triangle \text{x}=25$

Then,

$\triangle\text{x}=-2$

For $\text{x}=27$

$\text{y}=(27)^{\frac{1}{3}}=3$

Let:

$\text{dx}=\triangle \text{x}=-2$

Now, $\text{y}=(\text{x})^{\frac{1}{3}}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{3(\text{x})^{\frac{2}{3}}}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}=27}=\frac{1}{27}$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1}{27}\times(-2)=0.07407$

$\Rightarrow\triangle\text{y} =0.07407$

$\therefore(25)^{\frac{1}{3}}=\text{y}+\triangle\text{y}=2.9259$

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Question 394 Marks
Using differentials, find the approximate values of the following:
$(1.999)^5$
Answer
Consider the function $\text{y}=\text{f} (\text{x})=\text{x}^5$

Let:

$\text{x}=5$

$\text{x}+\triangle \text{x}=1.999$

Then,

$\triangle\text{x}=0.001$

For $\text{x}=2$

$\text{y}=2^5=32$

Let:

$\text{dx}=\triangle \text{x}=-0.001$

Now, $\text{y}=\text{x}^5$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=5\text{x}^4$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}=2}=80$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=80\times(-0.001)=-0.08$

$\Rightarrow\triangle\text{y} =-0.08$

$\therefore1.999^5=\text{y}+\triangle\text{y}=31.92$

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Question 404 Marks
If there is an error of 0.1% in the measurement of the radius of a sphere, find approximately the percentage error in the calculation of the volume of the sphere.
Answer
Let x be the radius of sphere,
$\triangle\text{x}=0.1\%\ \text{of}\ \text{x}$
$\triangle\text{x}=0.001\text{x}$
Now,
Let y = volume of sphere
$\text{y}=\frac{4}{3}\pi\text{x}^3$
$\frac{\text{dy}}{\text{dx}}=4\pi\text{x}^2$
$\triangle\text{y}=\Big(\frac{\text{dy}}{\text{dx}}\Big)\times\triangle\text{x}$
$=4\pi\text{x}^2\times0.001\text{x}$
$=\frac{4}{3}\pi\text{x}^3(0.003)$
$=\frac{0.3}{100}\times\text{y}$
$\triangle\text{y}=0.3\%\ \text{of}\ \text{y}$
so, percentage error in volume of error = 0.3%.
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Question 414 Marks
Using differentials, find the approximate values of the following:
$(0.007)^{\frac{1}{3}}$
Answer
Let $\text{x}=0.008,\\\text{x}+\triangle\text{x}=0.007$
$\triangle\text{x}=0.007-0.008$
$\triangle\text{x}=-0.001$
Let $\text{y}=\text{x}^{\frac{1}{3}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{3(\text{x})^{\frac{2}{3}}}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}\Rightarrow0.008}=\frac{1}{3(0.008)^{\frac{2}{3}}}$
$=\frac{100}{2}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}\Rightarrow0.008}=8.333$
$\triangle\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}\Rightarrow0.008}\times\triangle\text{x}$
$=(8.333)(-0.001)$
$\triangle\text{y}=-0.008333$
$(0.007)^{\frac{1}{3}}=\text{y}+\triangle\text{y}$
$=\text{x}^{\frac{1}{3}}-0.008333$
$=(0.008)^{\frac{1}{3}}-0.008333$
$=0.2-0.008333$
$(0.007)^{\frac{1}{3}}=0.191667$
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