Errors and Approximations MCQ Quiz - Objective Question with Answer for Errors and Approximations - Download Free PDF

Calculation

$81.5$ = $81+0.5$.

Let $f(x)=x^{\frac{1}{4}}$. 

$f(x+\mathrm{\Delta }x)\approx f(x)+f^{\prime }(x)\mathrm{\Delta }x$

let $x=81$ and $\mathrm{\Delta }x=0.5$.

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

$f^{\prime }(x)=\frac{1}{4}{x}^{\frac{1}{4}-1}=\frac{1}{4}{x}^{-\frac{3}{4}}=\frac{1}{4x^{\frac{3}{4}}}$

$f^{\prime }(81)=\frac{1}{4(81{)}^{\frac{3}{4}}}=\frac{1}{4(3^4{)}^{\frac{3}{4}}}=\frac{1}{4(3^3)}=\frac{1}{4(27)}=\frac{1}{108}$

$f(81.5)\approx f(81)+f^{\prime }(81)(0.5)$

$f(81.5)\approx 3+\frac{1}{108}(0.5)=3+\frac{0.5}{108}=3+\frac{1}{216}$

$f(81.5)\approx 3+0.0046296\approx 3.0046$

Hence option 3 is correct.

Calculation

Let $A=1+\alpha +{\alpha }^2+\dots$

Then $y=A{e}^{nx}$

Taking the natural logarithm of both sides:

⇒ $\ln y=\ln (A{e}^{nx})$

⇒ $\ln y=\ln A+\ln e^{nx}$

⇒ $\ln y=\ln A+nx$

Differentiating both sides with respect to $x$:

⇒ $\frac{1}{y}\frac{dy}{dx}=0+n$

⇒ $\frac{1}{y}\frac{dy}{dx}=n$

⇒ $\frac{dy}{y}=ndx$

The relative error in $y$ is $\frac{dy}{y}$.

The error in $x$ is $dx$.

Therefore, the relative error in $y$ is $n$ times the error in $x$.

⇒ Relative error in $y$ = $n\times$ (Error in $x$)

∴ The relative error in $y$ is $n$ times the error in $x$.

Hence option 3 is correct

Formula Used:

Surface area of sphere (A) = $4\pi r^2$

Volume of sphere (V) = $\frac{4}{3}\pi r^3$

Calculation:

Given:

Radius of the sphere (r) = 7 cm

Error in measuring surface area (ΔA) = 0.08 sq. cm

Differentiating the surface area formula, we get

⇒ $\frac{dA}{dr}=8\pi r$

Now, approximate error in radius (Δr) can be found using:

⇒ $\Delta A\approx \frac{dA}{dr}\Delta r$

⇒ $\Delta r\approx \frac{\Delta A}{dA/dr}=\frac{\Delta A}{8\pi r}$

⇒ $\Delta r\approx \frac{0.08}{8\times \pi \times 7}\approx \frac{0.01}{\pi \times 7}$ cm

Differentiating the volume formula, we get

⇒ $\frac{dV}{dr}=4\pi r^2$

⇒ $\Delta V\approx \frac{dV}{dr}\Delta r$

⇒ $\Delta V\approx (4\pi r^2)\Delta r$

⇒ Substituting the values:

⇒ $\Delta V\approx (4\times \pi \times 7^2)\times \frac{0.01}{\pi \times 7}$

⇒ $\Delta V\approx 0.28$ cubic cm

∴ The approximate error in volume is 0.28 cubic cm.

Hence option 1 is correct

Concept:

f(a + h) = f(a) + hf '(a)

Calculation:

Let f(x) = log₁₀x = $\frac{{\log }_{e}x}{{\log }_{e}10}$ = (log10 e)(log10 x) = 0.4343(log10 x)

⇒ f '(x) = $\frac{0.4343}{x}$

Let x = 998 = 1000 - 2 = a + h

⇒ a = 1000 and h = - 2

f(a) = f(1000) = log₁₀ 1000 = 3

f '(a) = f '(1000) = $\frac{0.4343}{1000}$ = 0.0004343

∴ f(998) = f(1000) - 2f '(1000)

⇒ f(998) = 3 - 2(0.0004343) = 2.9991314

⇒ log10 998 = 2.9991314

∴ The approximate value of log10 998 is 2.9991314.

The correct answer is Option 4.

Calculation

x = 45° and Δx = 2' = 0.035 

y = cotx

$\frac{dy}{dx}=-cose{c}^{2}x$

At x = 45° ⇒ $\frac{dy}{dx}=-2$

Δy = $\frac{dy}{dx}$ Δx 

⇒ Δy = - 0.07

cot 45°2' = y + Δy = cot45°- 0.07 = 1 - 0.07 = 0.93

Hence option 4 is correct

Concept: 

Let small charge in x be Δx and the corresponding change in y is Δy.

$\Delta \mathrm{y}={\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}\Delta \mathrm{x}={\mathrm{f}}^{\prime }(\mathrm{x})\Delta \mathrm{x}$

Now that Δy = f(x + Δx) - f(x)

Therefore, f(x + Δx) = f(x) + Δy

Calculation: 

Given:f(x) = 3x2 +  3

Let x + Δx = 3.01 = 3 + 0.01

Therefore, x = 3 and Δx = 0.01

f(x + Δx) = f(x) + Δy

= f(x + Δx) = f(x) + f'(x)Δx

= f(3.01) = 3x2 +  3 + (6x)Δx

= f(3.01) = 3(3)² + 3 + (6⋅3)(0.01)

= f(3.01) = 30 + 0.18

= f(3.01) = 30.18

Concept:

Let small charge in x be Δx and the corresponding change in y is Δy.

Therefore $\mathrm{\Delta }\mathrm{y}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\mathrm{\Delta }\mathrm{x}$

Calculation:

Given: y = 3x² + 2

To Find: total change in y = Δy

x changes from 10 to 10.1

So, Δx = 10.1 - 10 = 0.1

Now, y = 3x² + 2

Differentiating with respect to x, we get

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=6\mathrm{x}$

As we know, $\mathrm{\Delta }\mathrm{y}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\mathrm{\Delta }\mathrm{x}$

$\Rightarrow \mathrm{\Delta }\mathrm{y}=6\mathrm{x}\mathrm{\Delta }\mathrm{x}$

Put x = 10 and Δx = 0.1

$\therefore \mathrm{\Delta }\mathrm{y}=6\times 10\times 0.1=6\approx 6.03$

Concept: 

Let small charge in x be Δx and the corresponding change in y is Δy.

Therefore $\Delta \mathrm{y}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\Delta \mathrm{x}$

Calculation:

We have to find the value of $\sqrt{24.99}$

Let x + Δx = 24.99 = 25 - 0.01

Therefore, x = 25 and Δx = -0.01

Assume, $\mathrm{y}={\mathrm{x}}^{1/2}$          

Differentiating with respect to x, we get

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{1}{2}{\mathrm{x}}^{-1/2}=\frac{1}{2\sqrt{\mathrm{x}}}$

At x = 25

${\left[\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right]}_{\mathrm{x}=25}=\frac{1}{10}$ and y = $(25{)}^{1/2}=5$

As we know $\Delta \mathrm{y}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\Delta \mathrm{x}$

So, $\Delta \mathrm{y}=\frac{1}{10}\times (-0.01)=-0.001$

Therefore, approximate value of $\sqrt{24.99}=(24.99{)}^{1/2}$ = y + Δy = 5 - 0.001 = 4.999

Concept:

Applications of Calculus:

Approximation: f(x + Δx) ≈ f(x) + f'(x)Δx.

Calculation:

Let f(x) = √x be our function.

∴ f(x + Δx) ≈ √x + $\frac{1}{2}\left(\frac{1}{\sqrt{\mathrm{x}}}\right)$Δx.

Putting x = 4 and Δx = (2.718 - 4) = -1.282, we get:

⇒ f(4 - 1.282) ≈ √4 + $\frac{1}{2}\left(\frac{1}{\sqrt{4}}\right)$(-1.282)

⇒ √(2.718) ≈ 2 - $\frac{1}{2}\left(\frac{1}{2}\right)$(1.282)

⇒ √e ≈ 2 - 0.3205 = 1.6795.

Concept: 

Let small charge in x be Δx and the corresponding change in y is Δy.

Therefore $\Delta \mathrm{y}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\Delta \mathrm{x}$

Calculation:

We have to find an value of $\sqrt{4.24}$

Let x + Δx = 4.24 = 4 + 0.24

Therefore, x = 4 and Δx = 0.24

Assume, $\mathrm{y}={\mathrm{x}}^{1/2}$          

Differentiating with respect to x, we get

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{1}{2}{\mathrm{x}}^{-1/2}=\frac{1}{2\sqrt{\mathrm{x}}}$

At x = 4

${\left[\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right]}_{\mathrm{x}=4}=\frac{1}{2\sqrt{4}}=\frac{1}{4}$ and y = $(4{)}^{1/2}=2$

As we know $\Delta \mathrm{y}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\Delta \mathrm{x}$

So, $\Delta \mathrm{y}=\frac{1}{4}\times (0.24)=0.06$

Therefore, approximate value of $\sqrt{4.24}=(4.24{)}^{1/2}$ = y + Δy = 2 + 0.06 = 2.06

Concept:

Let the side of the cube is x meters,

Surface area of cube = S = 6 × (side)2

Approximate change in the surface area of the cube is given by: $\mathrm{\Delta }\mathrm{S}=\left(\frac{\mathrm{d}\mathrm{s}}{\mathrm{d}\mathrm{x}}\right)\mathrm{\Delta }\mathrm{x}$

Calculation:

Given: Side of the cube = x metres

Decrease in the side of cube = Δx = -2% of x            (Negative sign indicate decreasing the quantity)

⇒ Δx = -0.02x

Now, Surface area of a cube is given by,

 S = 6 × (side)² = 6x²

Differentiating with respect to x, we get

⇒ $\frac{\mathrm{d}\mathrm{S}}{\mathrm{d}\mathrm{x}}=12\mathrm{x}$

Now,

The approximate change in the surface area of a cube = $\mathrm{\Delta }\mathrm{S}=\left(\frac{\mathrm{d}\mathrm{s}}{\mathrm{d}\mathrm{x}}\right)\mathrm{\Delta }\mathrm{x}$

$\mathrm{\Delta }\mathrm{S}=\left(12\mathrm{x}\right)\times -0.02\mathrm{x}=-0.24{\mathrm{x}}^2$

Hence, the approximate change in the surface area of a cube is 0.24x² metres²

Concept: 

Let small charge in x be Δx and the corresponding change in y is Δy.

$\Delta \mathrm{y}={\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}\Delta \mathrm{x}={\mathrm{f}}^{\prime }(\mathrm{x})\Delta \mathrm{x}$

Now that Δy = f(x + Δx) - f(x)

Therefore, f(x + Δx) = f(x) + Δy

Calculation: 

Given: f (x) = 2x2 + x + 2.

f'(x) = 4x + 1

Let x + Δx = 2.002 = 2 + 0.002

Therefore, x = 2 and Δx = 0.002

f(x + Δx) = f(x) + Δy

= f(x + Δx) = f(x) + f'(x)Δx

= f(2.002) = 2x2 + x + 2 + (4x + 1)Δx

= f(2.002) = 2(2)² + 2 + 2 + [4⋅(2) + 1](0.002)

= f(2.002) = 12 + (9)(0.002)

= f(2.002) = 12 + 0.018

= f(2.002) = 12.018

Concept: 

Let small charge in x be Δx and the corresponding change in y is Δy.

$\Delta \mathrm{y}={\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}\Delta \mathrm{x}={\mathrm{f}}^{\prime }(\mathrm{x})\Delta \mathrm{x}$

Now that Δy = f(x + Δx) - f(x)

Therefore, f(x + Δx) = f(x) + Δy

Calculation: 

Given: f(x) = 2x3 + 5x

f'(x) = 6x² + 5

Let x + Δx = 2.05 = 2 + 0.05

Therefore, x = 2 and Δx = 0.05

f(x + Δx) = f(x) + Δy

= f(x + Δx) = f(x) + f'(x)Δx

= f(2.05) = 2x3 + 5x + (6x2 + 5)Δx

= f(2.05) = 2(2)³ + 5(2) + [6⋅(2)² + 5](0.05)

= f(2.05) = 16 + 10 + (24 + 5)(0.05)

= f(2.05) = 26 + (29)(0.05)

= f(2.05) = 26 + 1.45 = 27.45

Solution:

Second-order approximation of higher-order systems:
Certain higher-order systems can be approximated as second-order systems and can be characterized by the parameters in the preceding section.

The conditions are as follows:

• Higher-order poles are much farther into the left half of the s-plane than the dominant second-order pair of poles. The response that results from a higher-order pole does not appreciably change the transient response expected from the dominant second-order poles.
• Closed-loop zeros near the closed-loop second-order pole pair are nearly canceled by the close proximity of higher-order closed-loop poles.
• Closed-loop zeros not canceled by the close proximity of higher-order closed-loop poles are far removed from the closed-loop second-order pole pair.

Concept:

Let the radius of a sphere is r meters,

Volume of the sphere = V = $\frac{4}{3}\pi {\mathrm{r}}^3$

Approximate error in the volume of a sphere is given by:  $\mathrm{\Delta }\mathrm{V}=\left(\frac{\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{r}}\right)\mathrm{\Delta }\mathrm{r}$

Calculation:

Given: Radius of a sphere = 4 m and Δr = 0.01m

Now, the volume of the sphere is given by,

V = $\frac{4}{3}\pi {\mathrm{r}}^3$

Differentiating with respect to r, we get

$\Rightarrow \frac{\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{r}}=\frac{4\pi }{3}\times 3{\mathrm{r}}^2=4\pi {\mathrm{r}}^2$

Now,

The approximate error in the volume of sphere = $\mathrm{\Delta }\mathrm{V}=\left(\frac{\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{r}}\right)\mathrm{\Delta }\mathrm{r}$

$\therefore \mathrm{\Delta }\mathrm{V}=4\pi {\mathrm{r}}^2\times \mathrm{\Delta }\mathrm{r}$

⇒ ΔV = 4π × 4² × 0.01 = 0.64π m³

Hence, the approximate error in the volume of a sphere is 0.64π m³