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QUESTION: 1

The radius of air bubble is increasing at the rate of 0. 25 cm/s. At what rate the volume of the bubble is increasing when the radius is 1 cm.

Solution:

QUESTION: 2

The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 5x^{2} + 22x + 35^{. }Find the marginal revenue, when x = 7, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant

Solution:

QUESTION: 3

Find the approximate change in total surface area of a cube of side x metre caused by increase in side by 1%.

Solution:

QUESTION: 4

The volume of cube is increasing at the constant rate of 3 cm^{3}/s. Find the rate of change of edge of the cube when its edge is 5 cm.

Solution:

Let V be the instantaneous volume of the cube.

dV/dt = 3 cm^{3}/s

Let x be the side of the cube.

V = x^{3}

dV/dt = 3x^{2} * (dx/dt)

3 = 3*(5^{2})*(dx/dt)

So, dx/dt = 1/25 cm/s

QUESTION: 5

The total cost associated with the production of x units of a product is given by C(x) = 5x^{2} + 14x + 6. Find marginal cost when 5 units are produced

Solution:

C(x) = 5x^{2} + 14x + 6

Marginal Cost M(x) = C’(x)

M(x) = 10x +14

So, M(5) = 50 + 14

= 64 Rs

QUESTION: 6

At a distance of 120ft from the foot of a tower the elevation of its top is 60o. If the possible error in measuring the distance and elevation are 1 inch and 1 minute, find approximate error in calculated height.

Solution:

Tan(z) = ^{h}⁄_{x
}

QUESTION: 7

A stone is dropped into a quiet lake and waves move in circles at a speed of 2cm per second. At the instant, when the radius of the circular wave is 12 cm, how fast is the enclosed area changing ?

Solution:

Rate of increase of radius dr/dt = 2 cm/s

Area of circle A = πr^{2}

dA/dt = π*(2r)*(dr/dt)

= π*(24)*2

= 48π cm^{2}/s

Rate of increase of area is 48π cm^{2}/s (increasing as it is positive).

QUESTION: 8

Using approximation find the value of

Solution:

Let x=4, Δx=0.01

y=x^½ = 2

y+Δy = (x+ Δx)^½ = (4.01)^½

Δy = (dy/dx) * Δx

Δy = (x^(-1/2))/2 * Δx

Δy = (½)*(½) * 0.01

Δy = 0.25 * 0.01

Δy = 0.0025

So, (4.01)^½ = 2 + 0.0025 = 2.0025

QUESTION: 9

Find the approximate value of f(10.01) where f(x) = 5x^{2} +6x + 3

Solution:

f(x) = 5x^{2} +6x + 3

f(10.01) = 5*(10.01)^{2} + 6*(10.01) + 3

To find (10.01)^{2}

Let p=10, Δp=0.01

y=p2 = 100

y+Δy = (p+ Δp)^{2} = (10.01)^{2}

Δy = (dy/dp) * Δp

Δy = 2*p* Δx

Δy = 2*10* 0.01

Δy = 20 * 0.01

Δy = 0.2

So, (10.01)^{2} = y + Δy

= 100.2

So,

f(10.01) = 5*(100.2) + 6*(10.01) + 3

= 501 + 60.06 + 3

= 564.06

QUESTION: 10

Given a function y = f(x) . Let Δx be the very small change in the value of x , then the corresponding change in the value of y that is Δy is approximately given by

Solution:

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