12th NCERT Application of Derivatives Exercise6.4 Questions 9
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Question (1)

Using differentials, find the approximate value of each of the following up to 3 places of decimal.
(i)$\sqrt {25.3}$

Solution

To find the approximate value we use $f(x \pm \partial x) = f(x) \pm \partial xf'(x)$ where $\partial x$ is very very small compare to x.
$\sqrt {25.3}$ $Let\;f(x) = \sqrt x \Rightarrow f(25) = \sqrt {25} = 5$ $f'(x) = \frac{1}{{2\sqrt x }} \Rightarrow f'(25) = \frac{1}{{2\sqrt {25} }} = \frac{1}{{10}}$ $f(25.3) = f(25 + 0.3)$ $= f(25) + 0.3f'(25)$ $= 5 + 0.3\left( {\frac{1}{{10}}} \right)$ $\begin{array}{l} = 5 + 0.03\\ = 5.03\end{array}$ (ii) $\sqrt {49.5}$

Solution

$\sqrt {49.5}$ $Let\;f(x) = \sqrt x \Rightarrow f(49) = \sqrt {49} = 7$ $f'(x) = \frac{1}{{2\sqrt x }} \Rightarrow f'(49) = \frac{1}{{2\sqrt {49} }} = \frac{1}{{14}}$ $f(49.5) = f(49 + 0.5)$ $= f(49) + 0.5f'(49)$ $= 7 + 0.5\left( {\frac{1}{{14}}} \right)$ $\begin{array}{l} = 7 + 0.036\\ = 7.036\end{array}$ (iii) $\sqrt {0.6}$

Solution

$\sqrt {0.6}$ $Let\;f(x) = \sqrt x \Rightarrow f(1) = \sqrt {1} = 1$ $f'(x) = \frac{1}{{2\sqrt x }} \Rightarrow f'(1) = \frac{1}{{2\sqrt {1} }} = \frac{1}{{2}}$ $f(0.6) = f(1 - 0.4)$ $= f(1) - 0.4f'(1)$ $= 1 - 0.4\left( {\frac{1}{{2}}} \right)$ $\begin{array}{l} = 1 - 0.2\\ = 0.8\end{array}$ (iv) ( 0.009)1/3

Solution

( 0.009)1/3
0.009 = 0.008 + 0.001, as we can find the cube root of 0.008 as 0.2
$f(x) = {x^{1/3}}$ $f(0.008) = {(0.008)^{1/3}} = 0.2$ $f'(x) = \frac{1}{3}{x^{ - 2/3}} = \frac{1}{{3{x^{2/3}}}}$ $f'(0.008) = \frac{1}{{3{{(0.008)}^{2/3}}}} = \frac{1}{{3{{(0.2)}^2}}} = \frac{1}{{0.12}} = 8.33$ $f(0.009) = f(0.008 + 0.001)$ $= f(0.008) + 0.001f'(0.008)$ $= 0.2 + 0.001(8.33)$ $= 0.2 + 0.00833$ $= 0.20833 \approx 0.208$ (v) (0.999)1/10

Solution

$f(x) = {x^{\frac{1}{{10}}}}$ $f'(x) = \frac{1}{{10}}{x^{ - 9/10}} = \frac{1}{{10{x^{\frac{9}{{10}}}}}}$ $f(1) = {1^{1/10}} = 1$ $f'(1) = \frac{1}{{10{{(1)}^{\frac{9}{{10}}}}}} = \frac{1}{{10}} = 0.1$ $f(x \pm \partial x) = f(x) \pm \partial xf'(x)$ $f(0.999) = f(1 - 0.001)$ $= f(1) - 0.001f'(1)$ $= 1 - 0.001(0.1)$ $= 1 - 0.0001$ $= 0.9999$ (vi) (15)1/4

Solution

$f(x) = {x^{\frac{1}{4}}}$ $f'(x) = \frac{1}{4}{x^{ - 3/4}} = \frac{1}{{4{x^{\frac{3}{4}}}}}$ $f(16) = {16^{1/4}} = 2$ $f'(16) = \frac{1}{{4{{(16)}^{\frac{3}{4}}}}} = \frac{1}{{4{{({2^4})}^{3/4}}}} = \frac{1}{{4(8)}} = 0.03125$ $f(x \pm \partial x) = f(x) \pm \partial xf'(x)$ $f(15) = f(16 - 1)$ $= f(16) - 1f'(16)$ $= 2 - 1(0.03125)$ $= 2 - 0.03125$ $= 1.96875$ (vii) (26)1/3

Solution

$f(x) = {x^{\frac{1}{3}}}$ $f'(x) = \frac{1}{3}{x^{ - 2/3}} = \frac{1}{{3{x^{\frac{2}{3}}}}}$ $f(27) = {27^{1/3}} = 3$ $f'(27) = \frac{1}{{3{{(27)}^{\frac{2}{3}}}}} = \frac{1}{{3{{({3^3})}^{2/3}}}} = \frac{1}{{3(9)}} = 0.03703$ $f(x \pm \partial x) = f(x) \pm \partial xf'(x)$ $f(26) = f(27 - 1)$ $= f(27) - 1f'(27)$ $= 3 - 1(0.03703)$ $= 3 - 0.03703$ $= 2.96297$ (viii) (255)1/4

Solution

$f(x) = {x^{\frac{1}{4}}}$ $f'(x) = \frac{1}{4}{x^{ - 3/4}} = \frac{1}{{4{x^{\frac{3}{4}}}}}$ $f(256) = {256^{1/4}} = 4$ $f'(256) = \frac{1}{{4{{(256)}^{\frac{3}{4}}}}} = \frac{1}{{4{{({4^4})}^{3/4}}}} = \frac{1}{{4(64)}} = 0.00390625$ $f(x \pm \partial x) = f(x) \pm \partial xf'(x)$ $f(255) = f(256 - 1)$ $= f(256) - 1f'(256)$ $= 4 - 1(0.00390625)$ $= 4 - 0.00391$ $= 3.99609$ (ix) (82)1/4

Solution

$f(x) = {x^{\frac{1}{4}}}$ $f'(x) = \frac{1}{4}{x^{ - 3/4}} = \frac{1}{{4{x^{\frac{3}{4}}}}}$ $f(81) = {81^{1/4}} = 3$ $f'(81) = \frac{1}{{4{{(81)}^{\frac{3}{4}}}}} = \frac{1}{{4{{({3^4})}^{3/4}}}} = \frac{1}{{4(27)}} = 0.00925$ $f(x \pm \partial x) = f(x) \pm \partial xf'(x)$ $f(82) = f(81 + 1)$ $= f(81) + 1f'(81)$ $= 3 + 1(0.00925)$ $= 3 + 0.00925$ $= 3.00925$ (x)(401)1/2

Solution

$f(x) = {x^{\frac{1}{2}}}$ $f'(x) = \frac{1}{2}{x^{ - 1/2}} = \frac{1}{{2{x^{\frac{1}{2}}}}}$ $f(400) = {400^{1/2}} = 20$ $f'(400) = \frac{1}{{2{{(400)}^{\frac{1}{2}}}}} = \frac{1}{{2{{({{20}^2})}^{1/2}}}} = \frac{1}{{2(20)}} = 0.025$ $f(x \pm \partial x) = f(x) \pm \partial xf'(x)$ $f(401) = f(400 + 1)$ $= f(400) + 1f'(400)$ $= 20 + 1(0.025)$ $= 20 + 0.025$ $= 20.025$ (xi)(0.0037)1/2

Solution

$f(x) = {x^{\frac{1}{2}}}$ $f'(x) = \frac{1}{2}{x^{ - 1/2}} = \frac{1}{{2{x^{\frac{1}{2}}}}}$ $f(0.0036) = {0.0036^{1/2}} = 0.06$ $f'(0.0036) = \frac{1}{{2{{(0.0036)}^{\frac{1}{2}}}}} = \frac{1}{{2{{({{0.06}^2})}^{1/2}}}} = \frac{1}{{2(0.06)}} = 8.33$ $f(x \pm \partial x) = f(x) \pm \partial xf'(x)$ $f(0.0037) = f(0.0036 + 0.0001)$ $= f(0.0036) + 0.0001f'(0.0036)$ $= 0.06 + 0.0001(8.33)$ $= 0.06 + 0.000833$ $= 0.060833$ (xii) (26.57)1/3

Solution

$f(x) = {x^{\frac{1}{3}}}$ $f'(x) = \frac{1}{3}{x^{ - 2/3}} = \frac{1}{{3{x^{\frac{2}{3}}}}}$ $f(27) = {27^{1/3}} = 3$ $f'(27) = \frac{1}{{3{{(27)}^{\frac{2}{3}}}}} = \frac{1}{{3{{({3^3})}^{2/3}}}} = \frac{1}{{3(9)}} = 0.03703$ $f(x \pm \partial x) = f(x) \pm \partial xf'(x)$ $f(26.57) = f(27 - 0.43)$ $= f(27) - 0.43f'(27)$ $= 3 - 0.43(0.037)$ $= 3 - 0.01591$ $= 2.98409$ (xiii) (81.5)1/4

Solution

$f(x) = {x^{\frac{1}{4}}}$ $f'(x) = \frac{1}{4}{x^{ - 3/4}} = \frac{1}{{4{x^{\frac{3}{4}}}}}$ $f(81) = {81^{1/4}} = 3$ $f'(81) = \frac{1}{{4{{(81)}^{\frac{3}{4}}}}} = \frac{1}{{4{{({3^4})}^{3/4}}}} = \frac{1}{{4(27)}} = 0.00925$ $f(x \pm \partial x) = f(x) \pm \partial xf'(x)$ $f(81.5) = f(81 + 0.5)$ $= f(81) + 0.5f'(81)$ $= 3 + 0.5(0.00925)$ $= 3 + 0.004625$ $= 3.004625$ (xiv) (3.968)3/2

Solution

$f(x) = {x^{\frac{3}{2}}}$ $f'(x) = \frac{3}{2}{x^{1/2}} = \frac{{3{x^{1/2}}}}{2}$ $f(4) = {4^{3/2}} = 8$ $f'(4) = \frac{{3{{(4)}^{1/2}}}}{2} = \frac{{3(2)}}{2} = 3$ $f(x \pm \partial x) = f(x) \pm \partial xf'(x)$ $f(3.968) = f(4 - 0.032$ $= f(4) - 0.032f'(4)$ $= 8 - 0.032(3)$ $= 8 - 0.096$ $= 7.904$ (xv) (32.15)1/5

Solution

$f(x) = {x^{\frac{1}{5}}}$ $f'(x) = \frac{1}{5}{x^{ - 4/5}} = \frac{1}{{5{x^{\frac{4}{5}}}}}$ $f(32) = {32^{1/5}} = 2$ $f'(32) = \frac{1}{{5{{(32)}^{\frac{4}{5}}}}} = \frac{1}{{5{{({2^5})}^{4/5}}}} = \frac{1}{{5(16)}} = 0.0125$ $f(x \pm \partial x) = f(x) \pm \partial xf'(x)$ $f(32.15) = f(32 + 0.15)$ $= f(32) + 0.15f'(32)$ $= 2 + 0.15(0.0125)$ $= 2 + 0.001875$ $= 2.001875$

Question (2)

Find the approximate value of f(2.01), where f(x) = 4x2 + 5x + 2.

Solution

$\begin{array}{l}f(2.01) = f(2 + 0.01)\\ = f(2) + 0.01f'(2)\end{array}$ $f(x) = 4{x^2} + 5x + 2$ $f(2) = 4{(2)^2} + 5(2) + 2 = 16 + 10 + 2 = 28$ $f'(x) = 8x + 5$ $f'(2) = 8(2) + 5 = 16 + 5 = 21$ $\begin{array}{l}f(2.01) = f(2 + 0.01)\\ = f(2) + 0.01f'(2)\end{array}$ $= 28 + 0.01(21)$ $= 28 + 0.21$ $= 28.21$

Question (3)

Find the approximate value of f(5.001), where f(x) = x3 - 7x2 + 15.

Solution

$f(5.001) = f(5 + 0.001)$ $= f(5) + 0.001f'(5)$ $f(x) = {x^3} - 7{x^2} + 15$ $f(5) = {(5)^3} - 7{(5)^2} + 15 = 125 - 175 + 15 = - 35$ $f'(x) = 3{x^2} - 14x$ $f'(5) = 3{(5)^2} - 14(5) = 75 - 70 = 5$ $f(5.001) = f(5 + 0.001)$ $= f(5) + 0.001f'(5)$ $= - 35 + 0.001(5)$ $= - 35 + 0.005$ $= - 34.995$

Question (4)

Find the approximate change in the volume V of a cube of side x meter caused by increasing the side by 1%.

Solution

The volume of cube depends on the side of cube. If there is error in the side there will be change in the volume.
As the side of cube increases by 1%.
$\partial x = 1\% x = \frac{x}{{100}}m,$ $\partial v = \frac{{dv}}{{dx}}\partial x$ $\begin{array}{l}v = {x^3}\\Diff.w.r.t.x\end{array}$ $\frac{{dv}}{{dx}} = 3{x^2}$ $\partial v = \frac{{dv}}{{dx}}\partial x$ $\partial v = 3{x^2}\frac{x}{{100}}$ $\partial v = \frac{3}{{100}}{x^3}$ $\partial v = 0.03{x^3}{m^3}$

Question (5)

Find the approximate change in the surface area of cube of side x meter caused by decreasing the side by 1%.

Solution

The surface area of the cube is depends on the side x of the cube .
The side decreasing at the 1%.
$\partial x = 1\% x = \frac{x}{{100}}m,$ $\partial A = \frac{{dA}}{{dx}}\partial x$ $A = 6{x^2}$ $Diff.w.r.t.x,$ $\frac{{dA}}{{dx}} = 6(2x) = 12x$ $\partial A = \frac{{dA}}{{dx}}\partial x$ $\partial A = 12x\frac{x}{{100}}$ $\partial A = \frac{{12}}{{100}}{x^2}$ $\partial A = 0.12{x^2}{m^2}$

Question (6)

If the radius of a sphere is measured as 7m with an error of 0.02m, then find the approximate error in calculating its volume.

Solution

The radius of the circle = 7m. The error in radius = 0.02 m.
$\partial r = 0.02m,r = 7$ $\partial V = \frac{{dV}}{{dr}}\partial r$ $V = \frac{4}{3}\pi {r^3}$ $Diff.w.r.t.r,$ $\frac{{dV}}{{dr}} = \frac{4}{3}\pi (3{r^2}) = 4\pi {r^2}$ $\partial V = \frac{{dV}}{{dr}}\partial r$ $\partial V = 4\pi {r^2}(0.02)$ $\partial V = 4\pi {(7)^2}(0.02)$ $\partial V = 3.92\pi {m^3}$

Question (7)

If the radius of a sphere is measured as 9m with an error of 0.03m, then find the approximate error in calculating its surface area.

Solution

The radius of the circle = 9m. The error in radius = 0.03 m.
$\partial r = 0.03m,r = 9$ $\partial A = \frac{{dA}}{{dr}}\partial r$ $A = 4\pi {r^2}$ $Diff.w.r.t.r,$ $\frac{{dA}}{{dr}} = 4\pi (2r) = 8\pi r$ $\partial A = \frac{{dA}}{{dr}}\partial r$ $\partial A = 8\pi r(0.03)$ $\partial A = 8\pi (9)(0.03)$ $\partial A = 2.16\pi {m^2}$

Question (8)

If f(x) = 3x2 + 15x + 5, then the approximate value of f(3.02) is
(A) 47.66   (B) 57.66   (C) 67.66   (D) 77.66

Solution

$\begin{array}{l}f(3.02) = f(3 + 0.02)\\ = f(3) + 0.02f'(3)\end{array}$ $f(x) = 3{x^2} + 15x + 5$ $f(3) = 3{(3)^2} + 15(3) + 5 = 27 + 45 + 5 = 77$ $f(x) = 3{x^2} + 15x + 5$ $Diff.w.r.t.x,$ $f'(x) = 6x + 15$ $f'(x) = 6(3) + 15 = 18 + 15 = 33$ $\begin{array}{l}f(3.02) = f(3 + 0.02)\\ = f(3) + 0.02f'(3)\end{array}$ $= 77 + 0.02(33)$ $= 77 + 0.66$ $= 77.66$ So D is the correct option.

Question (9 )

The approximate change in the volume of a cube of side x meter caused by increasing the side by 3% is
(A) 0.06x3 m3   (B)0.6x3 m3   (C) 0.09 x3 m3   (D) 0.9 x3 m3

Solution

The side of cube be x and volume be V. $\partial x = 3\% x = 0.03x\;m,$ $\partial V = \frac{{dV}}{{dx}}\partial x$ $V = {x^3}$ $Diff.w.r.t.x,$ $\frac{{dV}}{{dx}} = 3{x^2}$ $\partial V = \frac{{dV}}{{dx}}\partial x$ $\partial V = 3{x^2}(0.03x)$ $\partial V = (0.09){x^3}{m^3}$ So C is the correct option.