Application of Derivatives Test

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Application of Derivatives Test - 23

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Question 1
1 / -0

If an error of $$k\%$$ is made in measuring the radius of a sphere, then percentage error in its volume is

A
$$k\%$$

B
$$3k\%$$

C
$$2k\%$$

D
$$\dfrac23k\%$$

Solution

Volume of sphere $$V=\dfrac { 4 }{ 3 } \pi { r }^{ 3 }$$
$$\displaystyle \dfrac{dV}{dr}=4\pi r^{2}$$

Given, percentage error in measuring radius =k%
$$\Rightarrow \displaystyle \dfrac { \Delta r }{ r } =\dfrac { k }{ 100 } $$

$$\Rightarrow \displaystyle { \Delta r }=\dfrac { kr }{ 100 } $$

Now, approximate error in measuring V$$\displaystyle =dV= (\dfrac{dV}{dr}){ \Delta r}$$

$$\displaystyle = \dfrac{k}{100} {4\pi r^{3}} = \dfrac{3k}{100}V =3k\%$$ of V

Percentage error in measuring $$V =3k\%$$
Question 2
1 / -0

If the ratio of base radius and height of a cone is 1:2 and percentage error in radius is $$\lambda$$ %, then the error in its volume is

A
$$\lambda$$ %

B
$$2\lambda$$%

C
$$3\lambda$$%

D
none of these

Solution

Volume of cone $$V=\dfrac { 1 }{ 3 } \pi { r }^{ 2 }h $$
Given, $$\displaystyle \dfrac{r}{h}=\dfrac{1}{2}$$
$$\Rightarrow V=\dfrac { 2 }{ 3 } \pi { r }^{ 3 } $$
$$\displaystyle \dfrac{dV}{dr}=2\pi r^{2}$$
Percentage error in measuring r $$=\lambda$$%
$$\Rightarrow \displaystyle \dfrac{\Delta r}{r}=\dfrac{\lambda}{100}$$
$$\Rightarrow \displaystyle \Delta r =\dfrac{\lambda r}{100}$$
Approximate error in V $$\displaystyle=dV=(\dfrac{dV}{dr}) \Delta r $$
$$\displaystyle=\dfrac{\lambda}{100} (2\pi r^{3})=\lambda\%$$ of V
Percentage error in $$V =\lambda\%$$
Question 3
1 / -0

If $$T=2\pi \sqrt {\dfrac {l}{g}}$$, then relative errors in T and l are in the ratio

A
$$1/2$$

B
$$2$$

C
$$1/2\pi$$

D
none of these

Solution

$$\displaystyle T=2\pi \sqrt {\dfrac {l}{g}}$$
$$\displaystyle \dfrac{dT}{dl}=\dfrac{\pi}{\sqrt{lg}}$$
Approximate error in T $$=\displaystyle dT=(\dfrac{dT}{dl})\Delta l$$
$$\displaystyle = \dfrac{\pi}{\sqrt{lg}}\Delta l$$
Relative error in T $$\displaystyle =\dfrac{dT}{T}=\dfrac{1}{2l} \Delta l$$

Approximate error in l $$=\displaystyle dl=(\dfrac{dl}{dT})\Delta T$$
$$\displaystyle = \dfrac{\sqrt{lg}}{\pi}\Delta T$$
Relative error in l i$$\displaystyle =\dfrac{dl}{l}=\dfrac{1}{\pi} \sqrt{\dfrac{g}{l}}\Delta T$$

Required ratio $$\displaystyle =\dfrac{\dfrac{1}{2l} \Delta l}{\dfrac{1}{\pi} \sqrt{\dfrac{g}{l}}\Delta T}$$
$$\displaystyle =\dfrac {\pi}{2\sqrt{lg}} \dfrac{\Delta l}{\Delta T}$$
$$\displaystyle =\dfrac{1}{2}$$
Question 4
1 / -0

If the percentage error in the edge of a cube is 1, then error in its volume is

A
$$1 \%$$

B
$$2 \%$$

C
$$3 \%$$

D
none of these

Solution

Volume of cube $$V=x^{3}$$
$$\Rightarrow \displaystyle \dfrac{dV}{dx}=3x^{2}$$
Percentage error in x is 1%.
$$\Rightarrow \displaystyle \dfrac{\Delta x}{x}=\dfrac{1}{100}$$
$$\Rightarrow \displaystyle \Delta x=\dfrac{x}{100}$$
Approximate error in V $$\displaystyle=dV=(\dfrac{dV}{dx}) \Delta x$$
$$\displaystyle = \dfrac{3{x}^{3}}{100}$$
Percentage error in V $$\displaystyle= \dfrac{dV}{V} $$
$$\displaystyle=\dfrac{3}{100}=3\%$$
Question 5
1 / -0

In a $$\Delta ABC$$ if sides a and b remain constant such that $$\alpha$$ is the error in C, then relative error in its area is

A
$$\alpha \cot C$$

B
$$\alpha \sin C$$

C
$$\alpha\tan C$$

D
$$\alpha\cos C$$

Solution

Area of triangle $$S\displaystyle =\dfrac {1}{2}ab \sin C$$
$$\displaystyle \Rightarrow \dfrac {dS}{dC}=\dfrac {1}{2}ab \cos C$$
Now, approximate error in S is $$\Delta S=\dfrac {dS}{dC}\Delta C$$
$$\displaystyle\Rightarrow \Delta S=\dfrac {1}{2}ab \cos C \alpha [\because \Delta C=\alpha]$$
$$\displaystyle\Rightarrow \dfrac {\Delta S}{S}=\dfrac {\dfrac {1}{2}ab \cos C}{\dfrac {1}{2}ab \sin C}\alpha=\alpha \cot C$$
Question 6
1 / -0

The circumference of a circle is measured as $$56$$ cm with an error $$0.02$$ cm. The percentage error in its area is

A
$$\dfrac {1}{7}$$

B
$$\dfrac {1}{28}$$

C
$$\dfrac {1}{14}$$

D
$$\dfrac {1}{56}$$

Solution

Circumference of circle $$C=2\pi r=56cm$$

$$\Rightarrow \displaystyle r=\frac{28}{\pi}$$

Also, $$\displaystyle \frac{dC}{dr}=2\pi$$

Area of circle $$A=\pi r^{2}$$

$$\displaystyle \frac{dA}{dr}=2\pi r$$

$$\Rightarrow\displaystyle \frac{dA}{dC}=r =\frac{28}{\pi}$$

Approximate error in A $$=\displaystyle dA=(\frac{dA}{dC})\Delta C$$

$$= r (0.02)$$

$$\Rightarrow\displaystyle \frac{dA}{A}= \frac{0.02}{\pi r}=\frac{1}{1400}$$

Percentage error in A is $$\displaystyle\frac{1}{14}$$%
Question 7
1 / -0

In a $$\Delta ABC$$ the sides b and c are given. If there is an error $$\Delta A$$ in measuring angle A, then the error $$\Delta a$$ in side a is given by

A
$$\dfrac {S}{2a}\Delta A$$

B
$$\dfrac {2S}{a}\Delta A$$

C
bc sin A $$\Delta A$$

D
none of these

Solution

In $$\triangle ABC$$ we have
$$\Rightarrow d\left( 2bc\cos { A } \right) =d\left( { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } \right) \\ \Rightarrow -2bc\sin { A } dA=-2ada\\ \Rightarrow bc\sin { A } dA=ada$$
$$\displaystyle \Rightarrow \frac { 2 }{ a } \left( \frac { 1 }{ 2 } bc\sin { B } \right) dA=da$$
$$\displaystyle \Rightarrow da=\frac { 2S }{ a } dA$$
$$\displaystyle \Rightarrow \triangle a=\frac { 2S }{ a } dA\\ \left[ \because dx\equiv \triangle a\quad and\quad dA=BA \right] $$
Question 8
1 / -0

If the percentage error in measuring the surface area of a sphere is $$\alpha$$ %, then the error in its volume is

A
$$\dfrac {3}{2}\alpha\%$$

B
$$\dfrac {2}{3}\alpha\%$$

C
$$3\alpha\%$$

D
none of these

Solution

Volume of sphere $$V=\dfrac{4}{3}\pi r^{3}$$
$$\displaystyle \dfrac{dV}{dr}=4\pi r^{2}$$
Surface area of sphere $$S=4\pi r^{2}$$
$$\displaystyle \dfrac{dS}{dr}=8\pi r$$
$$\displaystyle\Rightarrow \dfrac{dV}{dS}=\dfrac{r}{2}$$
Also, given percentage error in measuring S is $$\alpha$$%

$$\Rightarrow \displaystyle \dfrac{\Delta S}{S}=\dfrac{\alpha}{100}$$

$$\Rightarrow \displaystyle \Delta S=\dfrac{\alpha S}{100}=\dfrac{4\pi r^{2}\alpha}{100}$$

Approximate error in V is $$\displaystyle = dV= (\dfrac{dV}{dS}) \Delta S$$

$$\displaystyle = \dfrac {\alpha}{100} (2\pi r^{3})$$

Percentage error in V=$$\displaystyle \dfrac{dV}{V}$$

$$\displaystyle=\dfrac{3}{2}\alpha$$%
Question 9
1 / -0

If an error of $$1^o$$ is made in measuring the angle of a sector of radius $$30 \ cm$$, then the approximate error in its area is

A
$$450 cm^2$$

B
$$25\pi cm^2$$

C
$$2.5\pi cm^2$$

D
none of these

Solution

Area of sector $$\displaystyle A=\dfrac{\pi r^{2}\theta}{360}$$
Given $$r=30 cm, d{\theta}=1^{0}$$
Approximate error in A is $$\displaystyle=dA=(\dfrac{dA}{d\theta})\Delta \theta$$
$$\displaystyle = \dfrac{900\pi}{360} $$
$$\displaystyle \Rightarrow dA =2.5 \pi cm^{2}$$
Question 10
1 / -0

A line L is perpendicular to the curve $$\displaystyle y = \dfrac {x^2}{4} - 2$$ at its point P and passes through (10, -1). The coordinates of the point P are

A
(2, -1)

B
(6, 7)

C
(0, -2)

D
(4, 2)

Solution

Let point $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ on curve $$ y=\cfrac { { x }^{ 2 } }{ 4 } -2$$
Slope of tangent is $$\dfrac { dy }{ dx } =\dfrac { { x }_{ 1 } }{ 2 } $$
So slope of perpendicular line is $$-\dfrac { 2 }{ { x }_{ 1 } } $$ ...(1)
That perpendicular line also passes through $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$\left( 10,-1 \right) $$
Slope of line from these point is $$\dfrac { { y }_{ 1 }+1 }{ { x }_{ 1 }-10 } $$ ...(2)
Equating (1) and (2) and solving we get,
$${ x }_{ 1 }=4$$ and $${ y }_{ 1 }=2$$
Thus point is $$\left( 4,2 \right) $$
Hence, option 'D' is correct.

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 23

Result

Application of Derivatives Test - 23

Score

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SHARING IS CARING

Question 1

$$k\%$$

$$3k\%$$

$$2k\%$$

$$\dfrac23k\%$$

Solution

Question 2

$$\lambda$$ %

$$2\lambda$$%

$$3\lambda$$%

none of these

Solution

Question 3

$$1/2$$

$$2$$

$$1/2\pi$$

none of these

Solution

Question 4

$$1 \%$$

$$2 \%$$

$$3 \%$$

none of these

Solution

Question 5

$$\alpha \cot C$$

$$\alpha \sin C$$

$$\alpha\tan C$$

$$\alpha\cos C$$

Solution

Question 6

$$\dfrac {1}{7}$$

$$\dfrac {1}{28}$$

$$\dfrac {1}{14}$$

$$\dfrac {1}{56}$$

Solution

Question 7

$$\dfrac {S}{2a}\Delta A$$

$$\dfrac {2S}{a}\Delta A$$

bc sin A $$\Delta A$$

none of these

Solution

Question 8

$$\dfrac {3}{2}\alpha\%$$

$$\dfrac {2}{3}\alpha\%$$

$$3\alpha\%$$

none of these

Solution

Question 9

$$450 cm^2$$

$$25\pi cm^2$$

$$2.5\pi cm^2$$

none of these

Solution

Question 10

(2, -1)

(6, 7)

(0, -2)

(4, 2)

Solution

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