A volleyball player serves the ball which takes a parabolic path given by the equation

$h(t)= -\frac {7}{2} t^2 + \frac {13}{2} t + 1$

, where h(t) is the height of ball at any time t (in seconds), $(t \geq 0)$.

Based on the above information, answer the following questions :

(i) Is h(t) a continuous function ? Justify.

(ii) Find the time at which the height of the ball is maximum.

(i) Since this is polynomial function, it is continous

(ii) we have

$h(t)= -\frac {7}{2} t^2 + \frac {13}{2} t + 1$

$h'(t) = -7t+ \frac {13}{2}$

For maxima

$-7t+ \frac {13}{2}=0$

or t=13/14 sec

In order to set up a rain water harvesting system, a tank to collect rain water is to be dug. The tank should have a square base and a capacity of $250 \; m^3$. The cost of land is Rs 5,000 per square metre and cost of digging increases with depth and for the whole tank, it is Rs $40,000 h^2$, where h is the depth of the tank in metres. x is the side of the square base of the tank in metres

Based on the above information, answer the following questions :

(i) Find the total cost C of digging the tank in terms of x.

(ii) Find $\frac {dC}{dx}$

(iii)Find the value of x for which cost C is minimum

(iv)Check whether the cost function C(x) expressed in terms of x is increasing or not, where x > 0

Given

The volume of the tank = $250 \; m^3$

Now since this square base

$x^2 \times h = 250$

Solving for h:

$h = \frac {250}{ x^2}$

(i) Total cost C of digging the tank in terms of x:

Now, the cost of digging the tank increases with depth, and for the whole tank, it is $40,000 h^2$.

Cost of digging (C) = $40000 h^2$

Substituting the value of h from above

$C = 40000 \times \frac {62500}{x^4}$

$C = 25 \times 10^8 \frac {1}{x^4}$

(ii) Derivative dC/dx:

$\frac {dC}{dx} =\frac {d}{dx} (25 \times 10^8 \frac {1}{x^4})$

$= -10^10 \times \frac {1}{x^5}$

(iii) For Minima

$\frac {dC}{dx}=0$

$-10^10 \times \frac {1}{x^5}=0$

So there is no value of x that minimizes the cost. This means that the cost C keeps decreasing as x increases without bound.

So, there is no minimum cost for digging the tank; it will keep getting cheaper as the tank's side x increases.

(iv) Since derivative is negative , it is a decreasing function

Find the maximum and minimum values of the function given by

f(x) = 5 + sin 2x

Since $-1 \geq sin 2x \geq 1$

Maximum Value is 6 and Minimum value is 4

The median of the equilateral triangle is increasing at the rate $2 \sqrt 3 \; cm/s$ . Find the rate at which side is increasing

Median is given by

$M= \frac {\sqrt 3}{2} a$

Now

$\frac {dM}{dt} =\frac {\sqrt 3}{2} \frac {da}{dt}$

Now $\frac {dM}{dt}=2 \sqrt 3$

Therefore

$\frac {da}{dt}=4 $ cm/s

Sum of the two number is 5. if the sum of the cubes of the number is least, then find the sum of the squares

Let x and y be the two numbers

$x+ y=5$

Now

$C= x^3 + y^3 $

This can be written in one variable as

$C=x^3 + (5-x)^3 $

$\frac {dC}{dx} = 3x^2 - 3(5-x)^2$

Now for Minima

$\frac {dC}{dx}=0$

$3x^2 - 3(5-x)^2=0$

or x = 5/2

Now

$\frac {d^2C}{dx^2} = 6x + 6(5-x)$

At x=5/2

$\frac {d^2C}{dx^2} = 30$

Hence it is least

Now y = 5-x=5/2

Now

$x^2 + y^2 = 25/2$

A tap is connected to such a tank whose conical part is full of water. Water is dripping out from a tap at the bottom at the uniform rate of $2 \; cm^3/s$. The semi-vertical angle of the conical tank is 45°

On the basis of given information, answer the following questions :

(i) Find the volume of water in the tank in terms of its radius r.

(ii) Find rate of change of radius at an instant when $r = 2 \sqrt 2 \; cm$.

(iii) (a) Find the rate at which the wet surface of the conical tank is decreasing at an instant when radius r = $r = 2 \sqrt 2 \; cm$.

(iii) (b) Find the rate of change of height at height is 4 cm.

Show that the function $f(x) =\frac {16 sinx }{4+cos x} - x$ is strictly decreasing in $(\frac {\pi}{2}, \pi)$

If the radius of a circle is increasing at a rate of 2 cm/s, how fast is the area of the circle increasing when the radius is 5 cm?

Let the radius be \( r \) and the area \( A \). We know \( A = \pi r^2 \).

Differentiating both sides with respect to time \( t \), we get \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \).

Given \( \frac{dr}{dt} = 2 \) cm/s and \( r = 5 \) cm, we find

\( \frac{dA}{dt} = 2\pi \times 5 \times 2 = 20\pi \) cm^{2}/s.

Find the maximum value of the function \( f(x) = 2x^3 - 15x^2 + 36x + 1 \).

First, find the derivative:

\( f'(x) = 6x^2 - 30x + 36 \).
Set \( f'(x) = 0 \) to find critical points:

\( 6x^2 - 30x + 36 = 0 \).

Solving this, \( x = 2 \) or \( x = 3 \).

To determine the maxima, check the second derivative or use the first derivative test.

Here, \( f''(x) = 12x - 30 \). Since \( f''(2) > 0 \) and \( f''(3) < 0 \), the function has a maximum at \( x = 3 \).

The maximum value is \( f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 28 \).

Find the equation of the tangent to the curve \( y = x^2 - 4x + 3 \) at \( x = 2 \).

First, find the derivative: \( y' = 2x - 4 \).

At \( x = 2 \), the slope of the tangent is \( y'(2) = 2(2) - 4 = 0 \).

The y-coordinate at this point is \( y(2) = 2^2 - 4(2) + 3 = -1 \).

The equation of the tangent is \( y = mx + c \).

Since the slope \( m = 0 \) and it passes through \( (2, -1) \),

\( -1 = 0(2) + c \), so \( c = -1 \).

Thus, the equation is \( y = -1 \).

Determine the intervals where the function \( f(x) = x^3 - 3x^2 - 9x + 10 \) is increasing or decreasing.

First, find the derivative:

\( f'(x) = 3x^2 - 6x - 9 \).

Set \( f'(x) = 0 \) to find critical points:

\( 3x^2 - 6x - 9 = 0 \).

Solving this, we get \( x = -1 \) and \( x = 3 \).

Test intervals around these points to determine sign

The function is increasing on \( (-\infty, -1) \) and \( (3, \infty) \), and decreasing on \( (-1, 3) \).

Use differentials to approximate the value of \( \sqrt[3]{8.1} \).

Let \( y = x^{1/3} \).

At \( x = 8 \), \( y = 2 \).

The differential \( dy \) is given by \( dy = \frac{1}{3}x^{-2/3} dx \).

With \( dx = 0.1 \) (since we are approximating \( \sqrt[3]{8.1} \)), \( dy = \frac{1}{3} \cdot 8^{-2/3} \cdot 0.1 \approx 0.0125 \). Thus, \( \sqrt[3]{8.1} \approx 2 + 0.0125 = 2.0125 \).

Find the area of the greatest rectangle which can be incribed in a ellipse

$\frac {x^2}{a^2} + \frac {x^2}{b^2}=1$

Let ABCD is a rectangle incribed and 2p and 2q being the side as shown below

As (p,q) lies on the ellipse

$\frac {p^2}{a^2} + \frac {q^2}{b^2}=1$

or

$q^2=b^2(1-\frac {p^2}{a^2})$

Now Area of the rectangle is given by

$A = 2p \times 2q $

$A= 4pq$

or

$A^2 = 16 p^2 q^2$

Or

$A^2= 16 p^2 b^2(1-\frac {p^2}{a^2})$

$A^2=16b^2(p^2-\frac {p^4}{a^2})$

Now

$\frac {dA^2}{dp} = 16b^2(2p-\frac {4p^3}{a^2})$

for A to be max, $A^2$ is to max

So,

$\frac {dA^2}{dp} =0$

$16b^2(2p-\frac {4p^3}{a^2})=0$

or $p^2=a^2/2$

Now

$\frac {d^2A^2}{dp^2} = 16b^2(2-\frac {12p^2}{a^2})$

Substituting the values of p

$\frac {d^2A^2}{dp^2} < 0$

Hence

Area is maximum at this value

$A=4pq$

$=4 \frac {a}{\sqrt 2} \times b^2(1-\frac {1}{2})= 2ab$

Find the point P on the curve $y^2=4ax$ which is nearest to the point (11a,0)

Any point on the parabola is $(at^2 ,2at)$

Now distance is given by

$D^2= (at^2 -11a)^2 + (2at)^2$

$D^2 = a^2t^4 -18a^2t^2 + 121a^2$

Now D to be min, $L=D^2$ is to max

$L= a^2t^4 -18a^2t^2 + 121a^2$

$\frac {dL}{dt} = 4a^2t^3 - 36a^2 t $

For minimum of L

$\frac {dL}{dt}=0$

$4a^2t^3 - 36a^2 t=0 $

or t=0,3,-3

Now $\frac {d^2L}{dt^2} = 12a^2t^2 - 36a^2$

for t=0, $\frac {d^2L}{dt^2} < 0$, hence maxima

for t=3,-3 $\frac {d^2L}{dt^2} > 0$, hence minima

Hence we have two points where the distance is minimum, the cordinates are

(9a ,6a) and (9a ,-6a)

What is the slope of the tangent to the curve \( y = x^3 - 3x + 2 \) at the point where \( x = 2 \)?

(a) 9

(b) 10

(c) 11

(d) 12

- The slope of the tangent to \( y = x^3 - 3x + 2 \) is given by the derivative \( y' = 3x^2 - 3 \).

- At \( x = 2 \), \( y' = 3(2)^2 - 3 = 12 - 3 = 9 \).

If \( y = \sin x \), then the derivative \( \frac{dy}{dx} \) is:

(a) \( \cos x \)

(b) \( -\cos x \)

(c) \( \sin x \)

(d) \( -\sin x \)

The derivative of \( y = \sin x \) is \( \frac{dy}{dx} = \cos x \).

Answer is (a) \( \cos x \)

The derivative of \( f(x) = e^{2x} \) is:

(a) \( 2e^{2x} \)

(b) \( e^{2x} \)

(c) \( 2e^x \)

(d) \( e^x \)

The derivative of \( f(x) = e^{2x} \) is obtained by using the chain rule. It is \( f'(x) = 2e^{2x} \).

Answer is (a) \( 2e^{2x} \)

What is the equation of the normal to the curve \( y = x^2 \) at \( x = 1 \)?

(a) \( y + x = 0 \)

(b) \( y - x = 0 \)

(c) \( y - x = 2 \)

(d) \( y + x = 2 \)

The slope of the normal to the curve \( y = x^2 \) at \( x = 1 \) is the negative reciprocal of the derivative at that point.

The derivative \( y' = 2x \), so at \( x = 1 \), \( y' = 2 \). The slope of the normal is \( -1/2 \).

The equation of the normal is \( y - 1 = -\frac{1}{2}(x - 1) \), which simplifies to \( y + x = 2 \).

if \( y = \ln(x) \), the second derivative \( \frac{d^2y}{dx^2} \) is:

(a) \( -\frac{1}{x^2} \)

(b) \( \frac{1}{x^2} \)

(c) \( -\frac{1}{x} \)

(d) \( \frac{1}{x} \)

The second derivative of \( y = \ln(x) \) is \( \frac{d^2y}{dx^2} = -\frac{1}{x^2} \).

The maximum value of the function \( f(x) = -x^2 + 4x - 3 \) is:

(a) 4

(b) -1

(c) 1

(d) -4

To find the maximum value of \( f(x) = -x^2 + 4x - 3 \), first find the derivative and set it to zero to find critical points.

\( f'(x) = -2x + 4 \). Setting \( f'(x) = 0 \) gives \( x = 2 \).

The maximum value is \( f(2) = -(2)^2 + 4(2) - 3 = -4 + 8 - 3 = 1 \).

The point at which the function \( f(x) = x^3 - 3x^2 + 4 \) has a horizontal tangent is:

(a) \( x = 0 \)

(b) \( x = 1 \)

(c) \( x = 2 \)

(d) \( x = 3 \)

The function \( f(x) = x^3 - 3x^2 + 4 \) has a horizontal tangent when its derivative is zero.

\( f'(x) = 3x^2 - 6x \). Setting \( f'(x) = 0 \) gives \( x(3x - 6) = 0 \). Thus, \( x = 0 \) or \( x = 2 \).

The rate of change of the area of a circle with respect to its radius is:

(a) \( 2\pi r \)

(b) \( \pi r^2 \)

(c) \( \pi r \)

(d) \( 2\pi \)

- The rate of change of the area \( A \) of a circle with respect to its radius \( r \) is given by \( \frac{dA}{dr} \). Since \( A = \pi r^2 \), \( \frac{dA}{dr} = 2\pi r \).

The absolute minimum value of f(x)=2 sinx in $[0,\frac {3\pi}{2}]$ is

(a) 0

(b) 1

(c) -1

(d) -2

we know that

$-1 \leq sinx \leq 1$

Also sinx obtained its minimum value of -1 at $\frac {3\pi}{2}$

So absolute minimum value is -2

The function f(x)= x -sin x decreases for

(a) all x

(b) $0 < x < \frac {\pi}{4}$

(c) $x < \frac {\pi}{2}$

(d) No value of x

f(x)= x -sin x

f'(x) =1 -cos x

for decreasing

1 -cos x < 0

or 1< cos x

Which is not possible

Hence (d)

**Notes**-
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