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10 months ago

Application of Derivatives Maths

65. Using differentials, find the approximate value of each of the following up to 3 places of decimal.

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Vishal Baghel

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(i) Let y= ?x : Let x = 25 and x = 0. 3.

Then, ?y = ?x+?x $?x$

$=?25.3 ?25$

$=?25.3 ? 5$

$?25.3 = 5 + ? y ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? y ~ d y .$

= 5 + dy

= $5 + (\frac{d y}{d x}) A x .$

$= 5 + \frac{1}{2?x} ? x .$

$= 5 + \frac{1}{2?25} ? 0.3$

$= 5 + \frac{0.3}{2 * 5}$

= 5 + 0.03

$?25.3 = 5.030$

(ii) ?49.5

A.(ii)

Let y = ?x $.$ Let x = 49 and x = 0.5.

Then, $? y =?x+?x ?x .$

$=?49.5 ?49$

$?49.5 = 7 + ? y = 7 + {(\frac{d y}{d x})}_{}^{? x}$

$= 7 + \frac{1}{2?x} ? x$

$= 7 + \frac{1}{2?49} * (0.5)$

$= 7 + \frac{0.5}{14}$

= 7 + 0.0357.

$?49.5 = 7 ? 035$

(iii) ?0.6

A.(iii)

Let y = ?x $.$ Let x = 0.64 and ?x = 0.04.

Then, $? y =?x+Ax ?x$

$=?0.64-0.04 ?0.64 .$

$=?0.6 ? 0.8$

$?0.6= 0.8 + ? y = 0.8 + (\frac{d y}{d x}) ? x$

$= 0.8 + \frac{1}{2?x} ? x$

$= 0.8 + \frac{1}{2?0.64} * (? ? 0.04)$

$= 0.8 \frac{? 0.04}{2 * 0.8}$

= 0.8 - 0.025.

= 0.775.

(iv) ${(0.009)}^{\frac{1}{3}}$

A.(iv)

Let $y = x^{\frac{1}{3}}$ Let x = 0.008 and ?x = 0.00 1.

Then, ?y = ${[x + ? x]}^{\frac{1}{3}} ? {[x]}^{\frac{1}{3}}$

$= {(0.009)}^{\frac{1}{3}} ? {(0.008)}^{\frac{1}{3}}$

${(0.009)}^{\frac{1}{3}} = 0.2 + ? y = 0.2 + \frac{d y}{d x} ? ? x$

$= 0.2 + \frac{1}{3 {(x^{\frac{1}{3}})}^{2}} ? x$

$= 0.2 + \frac{1}{3 {[{(0.008)}^{\frac{1}{3}}]}^{2}} * (0.001)$

$= 0.2 + \frac{0.001}{3 * {(0.0)}^{2}} = 0.2 + \frac{0.001}{0.12}$

= 0.2 + 0.0083.

= 0.208.

(v) ${(0.999)}^{\frac{1}{10}}$

A.(v)

Let $y = x^{\frac{1}{10}} .$ Let x = 1 and ?x = -0.001

Then, $? y = {(x + ? x)}^{\frac{1}{10}} ? x^{\frac{1}{10}}$

$= {(0.999)}^{\frac{1}{10}} ? {(1)}^{\frac{1}{10}}$

$? 1 {(0.999)}^{\frac{1}{10}} = 1 + A y$

$= 1 + \frac{d y}{d x} ? x .$

$= 1 + \frac{1}{10 x^{\frac{9}{10}}} * ? x$

$= 1 ? \frac{0.001}{10 (. 1^{\frac{9}{10}})} = 1 ? \frac{0.001}{10} = 1 ? 0.0001$

= 0.999.

(vi) ${(15)}^{\frac{1}{4}}$

A.(vi)

Let $y = x^{\frac{1}{4}} .$ Then, x = 16 and ?x = 1.

Then, $? y = {(x + A x)}^{\frac{1}{4}} ? x^{\frac{1}{4}} .$

$= {(15)}^{\frac{1}{4}} ? {(16)}^{\frac{1}{4}}$

$? {(15)}^{\frac{1}{4}} = 2 + A y = 2 + \frac{d y}{d x} A x$

$= 2 + \frac{1}{4 x^{\frac{3}{4}}} ? (? 1)$

$= 4 ? \frac{1}{4 {(16)}^{\frac{3}{4}}}$

$= 2 ? \frac{1}{4 * 8}$

$= 4 ? \frac{1}{32}$

$= \frac{64 ? 1}{32} = \frac{63}{32} = 1.968$

(vii) ${(26)}^{\frac{1}{3}}$

A.(vii)

Let $y = x^{\frac{1}{3}} .$ Let x = 27 and ?x = 1.

Then, $? y = {(x + ? x)}^{\frac{1}{3}} ? x^{\frac{1}{3}}$

$= {(26)}^{\frac{1}{3}} ? {(27)}^{\frac{1}{3}}$

$? {(26)}^{\frac{1}{3}} = 3 + 4 y = 3 + \frac{d y}{d x} A x = 3 + \frac{1}{3 x^{\frac{2}{3}}} A x,$

$= 3 ? \frac{1}{3 {(2 F)}^{\frac{2}{3}}}$

$= 3 ? \frac{1}{27}$

$= \frac{81 ? 1}{27} = \frac{80}{27} = 2.962 .$

(viii) ${(255)}^{\frac{1}{4}}$

A.(viii)

Let $y = x^{\frac{1}{4}} .$ Let x = 256 and ?x = 1.

Then, $? y = {(x + ? x)}^{\frac{1}{4}} ? x^{\frac{1}{4}} .$

$= {(255)}^{\frac{1}{4}} ? {(256)}^{\frac{1}{4}}$

$? {(255)}^{\frac{1}{4}} = {(256)}^{\frac{1}{4}} + 4 y = 4 + \frac{d y}{d x} A x .$

$= 4 + \frac{1}{4 x^{\frac{3}{4}}} ? 4 x$

$= 4 \frac{? 1}{4 {(256)}^{\frac{3}{4}}}$

$= 4 ? \frac{1}{256}$

$= \frac{1024 ? 1}{256}$

$= \frac{1023}{256} = 3.996$

(ix) ${(82)}^{\frac{1}{4}}$

A.(ix)

...more

(i) Let y= ?x : Let x = 25 and x = 0. 3.

Then, ?y = ?x+?x $?x$

$=?25.3 ?25$

$=?25.3 ? 5$

$?25.3 = 5 + ? y ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? y ~ d y .$

= 5 + dy

= $5 + (\frac{d y}{d x}) A x .$

$= 5 + \frac{1}{2?x} ? x .$

$= 5 + \frac{1}{2?25} ? 0.3$

$= 5 + \frac{0.3}{2 * 5}$

= 5 + 0.03

$?25.3 = 5.030$

(ii) ?49.5

A.(ii)

Let y = ?x $.$ Let x = 49 and x = 0.5.

Then, $? y =?x+?x ?x .$

$=?49.5 ?49$

$?49.5 = 7 + ? y = 7 + {(\frac{d y}{d x})}_{}^{? x}$

$= 7 + \frac{1}{2?x} ? x$

$= 7 + \frac{1}{2?49} * (0.5)$

$= 7 + \frac{0.5}{14}$

= 7 + 0.0357.

$?49.5 = 7 ? 035$

(iii) ?0.6

A.(iii)

Let y = ?x $.$ Let x = 0.64 and ?x = 0.04.

Then, $? y =?x+Ax ?x$

$=?0.64-0.04 ?0.64 .$

$=?0.6 ? 0.8$

$?0.6= 0.8 + ? y = 0.8 + (\frac{d y}{d x}) ? x$

$= 0.8 + \frac{1}{2?x} ? x$

$= 0.8 + \frac{1}{2?0.64} * (? ? 0.04)$

$= 0.8 \frac{? 0.04}{2 * 0.8}$

= 0.8 - 0.025.

= 0.775.

(iv) ${(0.009)}^{\frac{1}{3}}$

A.(iv)

Let $y = x^{\frac{1}{3}}$ Let x = 0.008 and ?x = 0.00 1.

Then, ?y = ${[x + ? x]}^{\frac{1}{3}} ? {[x]}^{\frac{1}{3}}$

$= {(0.009)}^{\frac{1}{3}} ? {(0.008)}^{\frac{1}{3}}$

${(0.009)}^{\frac{1}{3}} = 0.2 + ? y = 0.2 + \frac{d y}{d x} ? ? x$

$= 0.2 + \frac{1}{3 {(x^{\frac{1}{3}})}^{2}} ? x$

$= 0.2 + \frac{1}{3 {[{(0.008)}^{\frac{1}{3}}]}^{2}} * (0.001)$

$= 0.2 + \frac{0.001}{3 * {(0.0)}^{2}} = 0.2 + \frac{0.001}{0.12}$

= 0.2 + 0.0083.

= 0.208.

(v) ${(0.999)}^{\frac{1}{10}}$

A.(v)

Let $y = x^{\frac{1}{10}} .$ Let x = 1 and ?x = -0.001

Then, $? y = {(x + ? x)}^{\frac{1}{10}} ? x^{\frac{1}{10}}$

$= {(0.999)}^{\frac{1}{10}} ? {(1)}^{\frac{1}{10}}$

$? 1 {(0.999)}^{\frac{1}{10}} = 1 + A y$

$= 1 + \frac{d y}{d x} ? x .$

$= 1 + \frac{1}{10 x^{\frac{9}{10}}} * ? x$

$= 1 ? \frac{0.001}{10 (. 1^{\frac{9}{10}})} = 1 ? \frac{0.001}{10} = 1 ? 0.0001$

= 0.999.

(vi) ${(15)}^{\frac{1}{4}}$

A.(vi)

Let $y = x^{\frac{1}{4}} .$ Then, x = 16 and ?x = 1.

Then, $? y = {(x + A x)}^{\frac{1}{4}} ? x^{\frac{1}{4}} .$

$= {(15)}^{\frac{1}{4}} ? {(16)}^{\frac{1}{4}}$

$? {(15)}^{\frac{1}{4}} = 2 + A y = 2 + \frac{d y}{d x} A x$

$= 2 + \frac{1}{4 x^{\frac{3}{4}}} ? (? 1)$

$= 4 ? \frac{1}{4 {(16)}^{\frac{3}{4}}}$

$= 2 ? \frac{1}{4 * 8}$

$= 4 ? \frac{1}{32}$

$= \frac{64 ? 1}{32} = \frac{63}{32} = 1.968$

(vii) ${(26)}^{\frac{1}{3}}$

A.(vii)

Let $y = x^{\frac{1}{3}} .$ Let x = 27 and ?x = 1.

Then, $? y = {(x + ? x)}^{\frac{1}{3}} ? x^{\frac{1}{3}}$

$= {(26)}^{\frac{1}{3}} ? {(27)}^{\frac{1}{3}}$

$? {(26)}^{\frac{1}{3}} = 3 + 4 y = 3 + \frac{d y}{d x} A x = 3 + \frac{1}{3 x^{\frac{2}{3}}} A x,$

$= 3 ? \frac{1}{3 {(2 F)}^{\frac{2}{3}}}$

$= 3 ? \frac{1}{27}$

$= \frac{81 ? 1}{27} = \frac{80}{27} = 2.962 .$

(viii) ${(255)}^{\frac{1}{4}}$

A.(viii)

Let $y = x^{\frac{1}{4}} .$ Let x = 256 and ?x = 1.

Then, $? y = {(x + ? x)}^{\frac{1}{4}} ? x^{\frac{1}{4}} .$

$= {(255)}^{\frac{1}{4}} ? {(256)}^{\frac{1}{4}}$

$? {(255)}^{\frac{1}{4}} = {(256)}^{\frac{1}{4}} + 4 y = 4 + \frac{d y}{d x} A x .$

$= 4 + \frac{1}{4 x^{\frac{3}{4}}} ? 4 x$

$= 4 \frac{? 1}{4 {(256)}^{\frac{3}{4}}}$

$= 4 ? \frac{1}{256}$

$= \frac{1024 ? 1}{256}$

$= \frac{1023}{256} = 3.996$

(ix) ${(82)}^{\frac{1}{4}}$

A.(ix)

Let $y = x^{\frac{1}{4}} .$ let x = 81 and ?x = 1.

Then, $? y = {(x + 4 x)}^{\frac{1}{4}} ? x^{\frac{1}{4}}$

$= {(82)}^{\frac{1}{4}} ? {(81)}^{\frac{1}{4}}$

$7 {(82)}^{\frac{1}{4}} = 3 + 4 y = 3 + \frac{d y}{d x} ? x$

$= 3 + \frac{1}{4 x^{\frac{3}{4}}} ? ? x$

$= 3 + \frac{1}{4 {(81)}^{\frac{3}{4}}}$

$= 3 + \frac{1}{4 * 3^{3}}$

$= 3 + \frac{1}{108} = \frac{324 + 1}{108} = \frac{325}{108} = 3.001 .$

(x) ${(401)}^{\frac{1}{2}}$

A.(x)

Let y = $x^{\frac{1}{2}}$ = ?x $.$ Let x = 400 and ?x = 1.

Then, $? y = {(x + ? x)}^{\frac{1}{2}} ? x^{\frac{1}{2}} .$

$? ? ? y = {(401)}^{\frac{1}{2}} ? {(400)}^{\frac{1}{2}}$

$? ? {(401)}^{\frac{1}{2}} = 20 ? + ? y = 20 + \frac{d y}{d x} ? x$

$= 20 + \frac{1}{2?x} ? x$

$= 20 + \frac{1}{2?400}$

$= 20 + \frac{1}{2 * 20}$

$= 20 + \frac{1}{40}$

$= \frac{800 + 1}{40} = \frac{801}{40} = 20.025 .$

(xi) ${(0.0037)}^{\frac{1}{2}}$

A.(xi)

Let $y = x^{\frac{1}{2}}$ Let x = 0.0036 and ?x = 0.0001

Then, $? y =?x+?x ?x$

$=?0.0037 ?0.0036$

$?0.0037 = 0.06 + D y = 0.06 + \frac{d y}{d x} ? x = 0.06 + \frac{x}{2?x}$

$= 0.06 + \frac{0.001}{2?0.0036}$

$= 0.06 + \frac{0.001}{2 * 0.06}$

$= 0.06 + \frac{0.001}{0.12 .}$

= 0.06 + 0.0008.

= 0.0608.

(xii) ${(26.57)}^{\frac{1}{3}}$

A.(xii)

Let $y = x^{\frac{1}{3}} .$ Let x = 27 and ?x = 0.43.

Then, $? y = {(x + A x)}^{\frac{1}{3}} ? {(x)}^{\frac{1}{3}}$

$= {(26.57)}^{\frac{1}{3}} ? {(27)}^{\frac{1}{3}}$

$? {(26.57)}^{\frac{1}{3}} = 3 + 4 y = 3 + \frac{d y}{d x} ? x = 3 + \frac{1}{3 x^{\frac{2}{3}}} ? ? x .$

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Non adherence will lead to removal of the comment Cancel Post V Vishal Baghel Contributor-Level 10 The given of the parabola is y^{2} = 4 a x slope of tangent is given by \frac{d}{d x} y^{2} = \frac{d}{d x} 4 a x \to 2 y \frac{d y}{d x} = 4 a \Rightarrow \frac{d y}{d x} = \frac{2 a}{y} . {\frac{d y}{d x} |}_{(a t^{2}, 2 a t)} = \frac{2 a}{2 a t} = \frac{1}{t} so, slope of normal so, slope of normal = \frac{- 1}{(\frac{1}{t})} = - t Hence eq n of tangent at point (a t^{2}, 2 a t) is y - 2 a t = \frac{1}{t} (x - a t^{2}) \to t y - 2 a t^{2} = x - a t^{2} \to x - t y - a t^{2} + 2 a t^{2} = 0 x - t y + a t^{2} = 0 And eq n of normal at point (a t^{2}, 2 a t) is (y - 2 a t) = (- t) (x - a t^{2}) \to y - 2 a t = - t x + a t^{3} \Rightarrow t x + y - 2 a t + a t^{3} = 0 \Rightarrow t x + y - a t (2 + t^{2}) = 0 00New Question 10 months ago B.N.M Institute of Technology Engineering Which are the BE specialisations offered at BNMIT? 0 Follower 1 View Follow Answer Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel Post N Nishtha Shukla Guide-Level 15 B.N.M Institute of Technology offers a four-year BE programme divided into eight semesters. The institute offers BE at the UG level in various specialisations. Candidates who have completed Class 12 in relevant subjects can apply for any of the below mentioned BE specialisations: Computer Science & Engineering Artificial Intelligence & Machine Learning Electronics & Communication Engineering Electrical & Electronics Engineering Information Science & Engineering Mechanical Engineering 00New Question 10 months ago Application of Derivatives Maths 58. Find the equation of the normals to the curve y = x 3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0. 0 Follower 2 Views Follow Answer Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel Post V Vishal Baghel Contributor-Level 10 The given eq n of the curve is y = x^{3} + 2 x + 6_____(1) slope of tangent to the curve, \frac{d y}{d x} = 3 x^{2} + 2 so, slope of normal to the curve = \frac{- 1}{3 x^{2} + 2} Now, the line x + 14 y + 4 = 0 \to y = - \frac{1}{14} x - \frac{4}{14} compared to y = m x + c gives slope of line = - \frac{1}{14} As the normal is parallel to the line \Rightarrow \frac{- 1}{3 x^{2} + 2} = - \frac{1}{14} \to 3 x^{2} + 2 = 14 \Rightarrow 3 x^{2} = 12 \to x^{2} = 4 \Rightarrow x = \pm 2 When x = 2, y = {(2)}^{3} + 2 (2) + 6 = 8 + 4 + 6 = 18 and when x = - 2, y = {(- 2)}^{3} + 2 (- 2) + 6 = - 8 - 4 + 6 = - 6 The point of contact of the normal are (2, 18) and (-2, -6) Hence the eq n of normal are y - 18 = - \frac{1}{14} (x - 2) and y - (- 6) = \frac{- 1}{14} [x - (- 2)] \Rightarrow 14 y - 252 = - x + 2 \Rightarrow y + 6 = \frac{- 1}{14} (x + 2) \Rightarrow x + 14 y - 254 = 0 \Rightarrow 14 y + 84 = - x - 2 \Rightarrow x + 14 y + 86 = 0 . 0036373839404142434445Register to get relevant Questions & Discussions on your feed Login or Register Ask & Answer Panel of Experts View Experts Panel .write-a-review-widget h2.tbSec2 {
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