Unit-1

Calculus

Question-1: Prove that the evolute of parabola is given by

Ans 1) It is given that-

If (X,Y) are the coordinates of the centre of curvature at any point P(x , y) on the curve y = f(x), then X and Y are given as-

…………….. (1)

Now consider the equation of parabola (given)

On differentiating w.r.t x-

Again differentiating w.r.t.x-

Put these derivatives in (1), we get-

Now consider X,

Here we get-

Now consider,

Here we get-

Now

Taking L.H.S of

Taking R.H.S of

Hence proved.

Question-2: Find the evolute of the ellipse .

Ans 2) It is given that-

The parametric equations are

Now,

and

So that-

Which gives,

Co-ordinates of centre of curvature are (X , Y),

…………….. (1)

Consider X,

We get-

Now consider Y,

So that we get-

Eliminating from X and Y, we get,

and

We know that

Which gives on solving-

Which is the required evolute.

Question-3: Evaluate.

Ans 3) Here we notice that f:x→cos x is a decreasing function on [a , b],

Therefore by the definition of the definite integrals-

Then

Now,

Here

Thus

Question-4: Evaluate

Ans 4) Here is an increasing function on [1 , 2]

So that,

…. (1)

We know that-

And

Then equation (1) becomes-

Question-5: Evaluate dx

Ans 5)

dx = dx

= γ(5/2)

= γ (3/2+ 1)

= 3/2 γ(3/2 )

= 3/2. ½ γ (½ )

= 3/2. ½ π

= ¾ π

Question-6: Show that

Ans 6)

=

=

= ) .......................

=

=

Question-7: Determine the area enclosed by the curves-

Ans 7) We know that the curves are equal at the points of interaction, thus equating the values of y of each curve-

Which gives-

By factorization,

Which means,

x = 2 and x = -3

By determining the intersection points the range the values of x has been found-

x | -3 | -2 | -1 | 0 | 1 | 2 |

1 | 10 | 5 | 2 | 1 | 2 | 5 |

And

x | -3 | 0 | 2 |

y = 7 - x | 10 | 7 | 5 |

We get the following figure by using above two tables-

Area of shaded region =

=

= ( 12 – 2 – 8/3 ) – (-18 – 9/2 + 9)

=

= 125/6 square unit

Question-8: Determine the area bounded by three straight lines y = 4 – x, y = 3x and 3y = x

Ans 8) We get the following figure by using the equations of three straight lines-

y = 4 – x, y = 3x and 3y = x

Area of shaded region-

Qustion-9: Find the area enclosed by the two functions-

and g(x) = 6 – x

Ans 9) We get the following figure by using these two equations

To find the intersection points of two functions f(x) and g(x)-

f(x) = g(x)

On factorizing, we get-

x = 6, -2

Now

Then, area under the curve-

A =

Therefore the area under the curve is 64/3 square unit.

Question-10: Find the volume of the solid of revolution formed by revolving R around y-axis of the function f(x) = 1/x over the interval [1 , 3].

Ans 10) The graph of the function f(x) = 1/x will look like-

The volume of the solid of revolution generated by revolving R(violet region) about the y-axis over the interval [1 , 3]

Then the volume of the solid will be-