Informative line

### Errors And Approximation

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

#### 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.

#### 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

# 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                $$dy=f'(x)dx$$

#### 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.

#### 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

# 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'

#### 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

# Evaluation of dy when values of dx and x are given

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

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

#### 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

# 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 =  $$\Delta V$$)

$$=4\pi\,a^2\,t$$

#### 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 $$\dfrac {dr}{r}×100$$

#### 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