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Errors And Approximation

Learn how to use linear approximation to approximate the value in calculus. Practice differentials & relative error & linear approximation examples in calculus.

Linear Approximations

Sometimes for a complicated function, we do not know or it is difficult to find the value at a particular value of x, in these cases we use the tangent line at\( (a,\, f(a))\) as an approximation to the curve \(y = f(x) \)where x is near 'a'.'.

Consider a curve \(y = f(x) \) and points \(A\, (a,\, f(a))\) and \(B\, (b,\, f(b))\) on it.

\(\therefore\) Slope of chord \(AB=\dfrac {f(b)-f(a)}{b-a}\)

If B is close to A or \(b-a\rightarrow0\), we can say that slope of tangent at 

\(A\cong\dfrac {f(b)-f(a)}{b-a}\)

  • Now let \(b=a+h\), where 'h' is small.

\(\therefore \;\dfrac {dy}{dx}\Bigg|_{x=a}\cong\dfrac {f(a+h)-f(a)}{h}\)

\(\therefore f(a+h)\cong f(a)+h\dfrac {dy}{dh}\Bigg|_{x=a}\)

  • We say that

\(L(x)=f(a) + (x-a) f'(a)\)                 (a + h = x)

is the linearization of 'f' at a.

  • We can approximate the values of \('f'\) at values of x close to \('a'\) by this approximation formula.

Steps for finding the linearization of 'f' at x = a

1. Find f '(x)

2. Use  \(L(x)=f(a) + (x-a) f'(a)\)

(a will be given)

  • \(L(x)\) gives the approximate value of function when x is close to 'a'

Illustration Questions

 Find the linearization \(L(x)\) of \(f(x)=x^{4/3}\) at \(a = 8\).

A \(\dfrac {7x}{4}-\dfrac {1}{5}\)

B \(\dfrac {7x}{6}+12\)

C \(\dfrac {8x}{3}-\dfrac {16}{3}\)

D \(\dfrac {8x}{5}+\dfrac {1}{4}\)

×

\(L(x)=f(a) + (x-a) f'(a)\)

Given     \(f(x)=x^{4/3}\)

\(\Rightarrow f'(x)=\dfrac {4}{3}x^{1/3}\)

\(L(x)=f(8)+(x-8)×f'(8)\)

\(=8^{4/3}+(x-8)×\dfrac {4}{3}×8^{1/3}\)

\(=16+(x-8)×\dfrac {4}{3}×2\)

\(=16+\dfrac {8}{3}x-\dfrac {64}{3}\)

\(\Rightarrow L(x)=\dfrac {8x}{3}-\dfrac {16}{3}\)

 Find the linearization \(L(x)\) of \(f(x)=x^{4/3}\) at \(a = 8\).

A

\(\dfrac {7x}{4}-\dfrac {1}{5}\)

.

B

\(\dfrac {7x}{6}+12\)

C

\(\dfrac {8x}{3}-\dfrac {16}{3}\)

D

\(\dfrac {8x}{5}+\dfrac {1}{4}\)

Option C is Correct

Illustration Questions

Use linear approximation to approximate the value of \((1.99)^3\).

A 10.23

B 7.67

C 8.43

D 7.88

×

Use \(f(x)=x^3\) and \(a = 2\)

\(L(x)=f(a)+(x-a).f'(a)\)

\( \Rightarrow L(x)=f(2)+(x-2).f'(2)\)

\( \Rightarrow L(x)=2^3+(x-2)×3×2^2\)

\( \Rightarrow L(x)=8+(x-2)×12\)

\( \Rightarrow L(x)=12x-16\)

\(L(1.99)=12×1.99-16=7.88\)

(Just put the value 1.99 in place of x)

Use linear approximation to approximate the value of \((1.99)^3\).

A

10.23

.

B

7.67

C

8.43

D

7.88

Option D is Correct

Differentials

We define differential \(dx\) as an independent variable such that differential dy is defined as 

\(dy=f'(x)\;dx\)    whenever  \(y=f(x)\).

  • \(dy\) is dependent variable which depends on \(dx\) and \(x\).
  • The idea of differentials is to find out the change in dependent variable \(dy\)  when there is a change in the independent variable \(dx\) for a certain function \(f(x).\)

\(\Delta y=\)Actual change in y

\(dy=\) amount by which tangent line rises or falls.

Illustration Questions

Find the differential \(dy\) for the function  \(y=\dfrac {\sqrt x}{1+x}\)

A \(\dfrac {1-x}{2\sqrt x{(1+x)}^2}\;dx\)

B \(\dfrac {1+x}{(1-x)^2}\;dx\)

C \(\dfrac {1-x}{2{(1+x)^2}}\;dx\)

D \(2\,tan\,x\;dx\)

×

The differential \(dy=f'(x)\;dx\)

\(f'(x)=\dfrac {d}{dx}\left (\dfrac {\sqrt x}{1+x}\right)=\dfrac {(1+x)\dfrac {d}{dx}\sqrt x-\sqrt x\dfrac {d}{dx}(1+x)}{(1+x)^2}^{\nearrow^{Quotient \,Rule}}\)

 

\(=\dfrac {(1+x)\dfrac {1}{2\sqrt x}-\sqrt x×1}{(1+x)^2}\)

\(=\dfrac {(1+x)-2x}{2\sqrt x(1+x)^2}\)

\(=\dfrac {1-x}{2\sqrt x(1+x)^2}\)

\(\therefore \; dy=\left (\dfrac {1-x}{2\sqrt x(1+x)^2}\right)\;dx\)

Find the differential \(dy\) for the function  \(y=\dfrac {\sqrt x}{1+x}\)

A

\(\dfrac {1-x}{2\sqrt x{(1+x)}^2}\;dx\)

.

B

\(\dfrac {1+x}{(1-x)^2}\;dx\)

C

\(\dfrac {1-x}{2{(1+x)^2}}\;dx\)

D

\(2\,tan\,x\;dx\)

Option A is Correct

Evaluation of dy when values of dx and x are given

The function \(y=f(x)\) can be evaluated for dy, using

\(dy=f'(x)\;dx\)

 

Illustration Questions

If \(y=\dfrac {x+3}{2x-1}\), find  \(dy\) when \(dx=0.1\) and \(x=1\)

A 0.6

B –0.7

C 1.2

D –1.4

×

\(dy=f'(x)\;dx\)

\(f'(x)=\dfrac {d}{dx}\left ( \dfrac {x+3}{2x-1} \right)\)

\(=\dfrac {(2x-1)\dfrac {d}{dx}(x+3)-(x+3)\dfrac {d}{dx}(2x-1)} {(2x-1)^2}\)

\(=\dfrac {(2x-1)-(x+3)×2}{(2x-1)^2}=\dfrac {-7}{(2x-1)^2}\)

\(\therefore \,dy=f'(x)\;dx\rightarrow\) Put x = 1, dx = 0.1

\(\Rightarrow \,dy=f'(1)×0.1\)

\(=\dfrac {-7}{(2-1)^2}×0.1=-0.7\)

\(\therefore \,dy=-0.7\)

If \(y=\dfrac {x+3}{2x-1}\), find  \(dy\) when \(dx=0.1\) and \(x=1\)

A

0.6

.

B

–0.7

C

1.2

D

–1.4

Option B is Correct

Difference between dx and delta x (\(\Delta x\)) and dy and delta y (\(\Delta y\) )

Consider a curve \(y=f(x)\)

A point \(P(x,f(x))\) and \(Q(x+\Delta x,\;f(x+\Delta x))\)are take on it.

We see that          \(\Delta y = f(x+\Delta x)-f(x)\)

whereas                \(dy=f'(x)dx\)

Illustration Questions

Find  \(\Delta y\) and \(dy\) for \(x=2\), \(dx=\Delta x=0.1\) and \(y=\dfrac {1}{x^2}\)

A \(dy=-0.5 \\ \Delta y=-0.48\)

B \(dy=-0.025 \\ \Delta y=-0.023\)

C \(dy=0.1 \\ \Delta y=0.09\)

D \(dy=-1.2\\ \Delta y=-1.3\)

×

\(\Delta y=f(x+\Delta x)-f(x)\)

Put \(x=2\)\(\Delta x=0.1\)

\(\Rightarrow \Delta y=f(2+0.1)-f(2)\)

\(=f(2.1)-f(2)\)

\(\therefore \Delta y=\dfrac {1}{(2.1)^2}-\dfrac {1}{2^2}=0.22676-0.25=-0.0233\)

\(dy=f'(x)\,dx\)

\(dy=\dfrac {-2}{x^3}\,dx\)

\(dy=\dfrac {-2}{8}×0.1\)

\(= \dfrac {-0.1}{4}\)

\(=-0.025\)

Find  \(\Delta y\) and \(dy\) for \(x=2\), \(dx=\Delta x=0.1\) and \(y=\dfrac {1}{x^2}\)

A

\(dy=-0.5 \\ \Delta y=-0.48\)

.

B

\(dy=-0.025 \\ \Delta y=-0.023\)

C

\(dy=0.1 \\ \Delta y=0.09\)

D

\(dy=-1.2\\ \Delta y=-1.3\)

Option B is Correct

Use of Differentials in Estimating the Errors

Suppose volume of sphere

\(V=\dfrac {4}{3}\pi r^3\), then if there is an error of \(\Delta r\) or \(dr\) in measurement of radius then the maximum error in the measurement of volume is 

\(d V=4\,\pi\,r^2\;dr\)

  • Relative Error = \(\dfrac {d V}{V}=\dfrac {4\,\pi\,r^2\;dr}{4/3\,\pi\,r^3}=\dfrac {3}{r}\;dr\)
  • Whenever two quantities are related an error in measurement of one quantity will lead to an error in measurement of other. Differentiate both sides of equation between two quantities.

Illustration Questions

The length of edge of cube is 20 cm, with possible error of 0.2 cm, estimate the maximum possible error in the volume of cube.

A 240 cm3

B 510  cm3

C 105  cm3

D 70 cm3

×

\(V =\ell^3\)

\(V=\) volume of cube, \(\ell=\) length of edge of cube

\(dV=3\ell^2\,dl\)

\(\ell=20\,cm,\;dl=0.2\)

\(dV=3×20^2×0.2\)

\(=3×400×0.2\)

\(=240\,cm^3\)

The length of edge of cube is 20 cm, with possible error of 0.2 cm, estimate the maximum possible error in the volume of cube.

A

240 cm3

.

B

510  cm3

C

105  cm3

D

70 cm3

Option A is Correct

Relative Error

Relative Error = \(\dfrac {\text {Maximum Error}}{\text {Total quantity}\nearrow}(\text {whose error is measured})\)

Suppose the radius of sphere is measured and found to be 'a' with maximum positive error of say 't' then 

\(V=\dfrac {4}{3}\,\pi\,r^3\)

\(\Rightarrow dV =4\pi\,\underbrace{r^2}_{a^2}\;\underbrace{dr}_{t}\)      ( \(dV \rightarrow\) approximate to error in volume =  \(\Delta V\))

\(=4\pi\,a^2\,t\)

Illustration Questions

The edge of cube is 20 cm with a possible error in measurement as 0.2 cm, find the relative error in measurement of surface area of cube.

A 0.02

B 0.04

C 0.03

D 0.09

×

\(S=6\,\ell^2\), where 

\(S =\) Surface area, \(\ell =\)length of edge

\(dS=12\,\ell\;d\ell\)

Put \(\ell=20\,cm\;d\ell=0.2\,cm\)

\(dS=12×20×0.2=48\,cm^2\)

Relative error = \(\dfrac {dS}{S}=\dfrac {48}{6×20^2}=0.02\,\)

The edge of cube is 20 cm with a possible error in measurement as 0.2 cm, find the relative error in measurement of surface area of cube.

A

0.02

.

B

0.04

C

0.03

D

0.09

Option A is Correct

Percentage Error

  • If errors are expressed in percentage, such errors are called percentage errors.
  • For example

Relative Error = \(\dfrac {\text {Measured Value - True Value}}{\text {True Value}}\)

  • Percentage Error = Relative error × 100
  • Suppose the radius of the sphere is \(r\) then its volume is \(V=\dfrac {4}{3}\,\pi\,r^3\)
  • If the error in the measured value of  \(r\) is denoted by \(dr\).
  • The relative error in the radius is given by \(\dfrac {dr}{r}\)
  • Percentage error will be given by \(\dfrac {dr}{r}×100\)

Illustration Questions

If the radius of sphere is given by 10 cm with a maximum possible error 0.01 cm. Determine the percentage error in measurement of volume of the sphere.

A 10 %

B 0.3%

C 0.7%

D 0.5%

×

 Volume of  sphere, \(V=\dfrac {4}{3}\,\pi \;r^3\)

Differential \(dV=4\,\pi\,r^2\;dr\)

Relative error in volume = \(\dfrac {dV}{V}\)

 \(=\dfrac {4\pi \,r^2\;dr}{\dfrac {4}{3}\,\pi\,r^3}\)

\(=3\;\dfrac {dr}{r}\)

\(3×\dfrac {0.01}{10}=0.003\)

 

\(Percentage\; error = Relative\; error\; \times\; 100\)

\(=\;0.003\;\times\;100\)

\(=\;0.3\text{%}\)

If the radius of sphere is given by 10 cm with a maximum possible error 0.01 cm. Determine the percentage error in measurement of volume of the sphere.

A

10 %

.

B

0.3%

C

0.7%

D

0.5%

Option B is Correct

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