B

Roll No.:Subject Code: BS-111Subject: Engineering Mathematics-IDated: 02nd Dec, 2008[9.00AM-12.00 Noon]Class: B.Tech./B.Arch. (1st Semester)Time : 20 min. M. Marks : 10.Note : Attempt All questions. Return this original sheet within 20 minutes. No overwriting, no cutting is allowed. 0.5 marks will be deducted for indicated incorrect response of each question. No deduction from the total score will be made if no response is indicated. Use blue/black Ball Pen only. Use of pencil is strictly prohibited. Exchange of calculators is not allowed.Q.No.Select the correct answer(s) in each of the following questions:

1.

If has radius of convergence is R (assumed finite), then radius of convergence of is

(a) R (b) (c) (c) None of these.

2.

If , , , then (a) 1 (b) 2 (c) 3 (d) 4

3. In double integrals, the total volume of the solid formed by the revolution of the area A about x-axis is evaluated by using the formula

(a) (b) (c) (d) None of these

4.

The plane containing unit vector (binormal) and unit tangent vector is known as (a) Osculating plane (b) Normal plane (c) Rectifying plane (d) None of these

5. An irrotational field is characterized by the following conditions:Write down the correct options:

(a) . (b)

(c) . (d) , if the domain is simply connected.

(e) Circulation along every closed curve is zero.

(f) Flux across every closed surface is zero.

Q.No.12345

Ans.

Fill in the blanks:

6. The total number of terms needed to compute the sum S of the geometric series with an error less than 0.01, when r = 0.9 are ..

7.

If unit tangent vector at a point to the curve C is and is the unit normal vector at that point then .

8.

If has direction of , k and are curvature and torsion, respectively, then .

9.

The vector normal to the surface , at the point is

State weather the following statement is true or false:

10.

If is the Jacobian of u, v, with respect to x, y and is the Jacobian of x, y, with respect to u, v, then .

Roll No.:Subject Code: BS-111Semester-1st Subject: Engineering Mathematics-IDated: 02nd Dec, 2008[9.00AM-12.00 Noon]Time: 2 Hours 40 min. Maximum Marks: 40Note : Attempt All questions. Marks are given against each question. Cross all the blank pages and blank portions of the answer book. No extra sheet will be supplied. Exchange of calculators is not allowed.Section A (Infinite Series)Q.No.1. State and prove Leibnitzs rule. (3)Q.No.2. Examine the behaviour of the infinite series

. (3)Section B (Differential Calculus)

Q.No.3. If . Show that . (3)

Q.No.4. Evaluate . (3)Q.No.5. A wire of length b is cut into two parts, which are bent in the form of a square and circle respectively. Find the least value of the sum of the areas so found by using Lagranges method. (3)Section C (Integral Calculus)Q.No.6. By single integral, find the volume and surface generated by the revolution of the

Astroid, , about the x-axis. (3)

Q.No.7. Using the concept of double integrals, evaluate , where R is the region

bounded by parallelogram , , , . (3)

Q.No.8. Find, by triple integration, the volume bounded above, by the sphere

and bounded below by the paraboloid . (4)

Section D (Vector Calculus)Q.No.9. (a) Discuss the physical interpretation of gradient of a scalar point function.

Also write a unit normal vector of a surface z = g(x, y). (2.5+0.5)

(b) Using the concept of surface integral, evaluate,

where and S is the triangular surface lying in the first octant. (4)

Q.No.10. (a) Show that may be written in the form

,where w(x, y) be a function which is continuous and has continuous first and second order partial derivatives in a domain of the xy-plane containing a region E of the type indicated in Greens theorem. Here s is the arc length of C. (4)

(b) State divergence theorem and hence evaluate

where and S is the upper part of the sphere

above XOY plane. (1+3)*** *** *** *** ****** *** ******

Roll No.:Subject Code: BS-111Subject: Engineering Mathematics-IDated: 02nd Dec, 2008[9.00AM-12.00 Noon]Class: B.Tech./B.Arch. (1st Semester)Time : 20 min. M. Marks : 10.Note : Attempt All questions. Return this original sheet within 20 minutes. No overwriting, no cutting is allowed. 0.5 marks will be deducted for indicated incorrect response of each question. No deduction from the total score will be made if no response is indicated. Use blue/black Ball Pen only. Use of pencil is strictly prohibited. Exchange of calculators is not allowed.Q.No.Select the correct answer(s) in each of the following questions:

1.

If has radius of convergence is R (assumed finite), then radius of convergence of is

(a) R (b) (c) (c) None of these.

2.

If , , , then (a) 1 (b) 2 (c) 3 (d) 4

3. In double integrals, the total volume of the solid formed by the revolution of the area A about x-axis is evaluated by using the formula

(a) (b) (c) (d) None of these

4.

The plane containing unit vector (binormal) and unit normal vector is known as (a) Osculating plane (b) Normal plane (c) Rectifying plane (d) None of these

5. An irrotational field is characterized by the following conditions:Write down the correct options:

(a) . (b)

(c) . (d) , if the domain is simply connected.

(e) Circulation along every closed curve is zero.

(f) Flux across every closed surface is zero.

Q.No.12345

Ans.bdcba d e

Fill in the blanks:

6. The total number of terms needed to compute the sum S of the geometric series with an error less than 0.01, when r = 0.9 are 66.

7.

If unit tangent vector at a point to the curve C is and is the unit normal vector at that point then .

8.

If has direction of , k and are curvature and torsion, respectively, then .

9.

The vector normal to the surface , at the point is

State weather the following statement is true or false:

10.

If is the Jacobian of u, v, with respect to x, y and is the Jacobian of x, y, with respect to u, v, then . True

Roll No.:Subject Code: BS-111Semester-1st Subject: Engineering Mathematics-IDated: 02nd Dec, 2008[9.00AM-12.00 Noon]Time: 3 Hours Maximum Marks: 50Note : Attempt All questions. Marks are given against each question.

*** *** *** *** ****** *** ******

Q.No.1. State and prove Leibnitzs rule. (3)

Statement: An alternating series converges if (i) each term is numerically less than its preceding term,

i.e. Mathematically for , and

(ii) .

If is oscillatory.

Proof: The given series is .Given (i) each term is numerically less than its preceding term,

i.e. (1)

(ii) . (2)Now consider the sum of 2n terms. It can be written as

. (3)

Also . (4)

Also [by (1)] (5)Since the expressions within the brackets in (3) and (4) are all positive. [by (1)]

Therefore, (3) The sequence is positive,

(4) The sequence is bounded above and always remains less than .

Also (5) The sequence is monotonically increasing.

Since we know, every monotonically increasing sequence, which is bounded above, converges. Therefore, the sequence converges.

Let us suppose (finite). (6)Thus, given alternating series is convergent, if we consider initially sum of even terms.To show: The uniqueness of this limit

Since . [by (2) and (6)]

This shows that tends to the same limit, whether n is even or odd.Hence the given series is convergent.To show: The oscillatory case

If , then .

The given series is oscillatory. This completes the proof.Q.No.2. Examine the behaviour of the infinite series

. (3)

Sol.: Here and so .

.

Case 1: When , then .

Hence is convergent (by DAlemberts ratio test).

Case 2: When , then

.

Hence is convergent (by DAlemberts ratio test).

Case 3: When , then

. In this case series becomes

This series is geometric series whose first term is and common ratio is unity. Hence by geometric series test the series is divergent when.

Section B (Differential Calculus)

Q.No.3. If . Show that . (3)

Sol.: Here

(i)Differentiating (i) partially w.r.t. x, we get

(ii)Differentiating (ii) partially w.r.t. x, we get

(iii)

Similarly, (iv)

Consider

. Ans.

Q.No.4. Evaluate . (3)

Sol.:

. Ans.

Q.No.5. A wire of length b is cut into two parts, which are bent in the form of a square and circle respectively. Find the least value of the sum of the areas so found by using Lagranges method. (3)Sol.: Let x and y be two parts into which the given wire is cut so that x + y = b.

Suppose the piece of wire of length x is bent into a square so that each side is and thus the area of the square is .=.Suppose the wire of length y is bent into a circle with perimeter y. So the area of this circle so formed is

. Since .Thus to find the minimum of the sum of the two areas subject to the constraint that sum x +y = b.

Let , where is Lagranges's multipliers.

.

Then

Solving, we get , .Substituting these values in the constraint condition x +y = b, we get

.

Thus and .Thus the least value of the sum of the areas of the square and circle is

. Ans.

Section C (Integral Calculus)Q.No.6. By single integral, find the volume and surface generated by the revolution of the

Astroid, , about the x-axis. (3) Sol.:

Bx-axisy-axis(a, 0)(0, a)

OCA

D

If an arc ABC rotated about x-axis, then we get required volume generated by this rotation.

Thus, by symmetry, the required volume generated = 2 volume generated by the curve lying in first quadrant

Volume generated

cubic units. Ans.

As we know, surface of revolution = .

Surface generated

sq. units. Ans.

Q.No.7. Using the concept of double integrals, evaluate , where R is the region

bounded by parallelogram , , , . (3)

Sol.: By changing the variables x, y to the new variables u, v, by the substitution (transformation) , , then the region R, i. e. parallelogram ABCD in the xy-plane becomes the region , i. e. rectangle RSPQ in the uv-plane, as shown in the figure, by taking , . (i)

Ou-axisv-axisv = 3u = 2v = 0RSPQOx-axisy-axisx + y = 03x - 2y = 33x 2y = 0x + y = 2ABCD

From (i), we have , .

.

Thus .

Since, and . Thus u varies from 0 to 2.

Also since , . Thus v varies from 0 to 3.

Thus the given integral in terms of new variables u, v is

. Ans.

Q.No.8. Find, by triple integration, the volume bounded above, by the sphere

and bounded below by the paraboloid . (4)

Sol.: Equation of the given sphere is and equation of the given paraboloid is

i.e. z varies from to .

Now

Since we have to find volume bounded above by the sphere and below the paraboloid . Thus z = -2a (rejected).

Thus equation of circle becomes

and y varies from to and similarly x varies from to

Required volume

Put , we get

Required volume

. Cubic units.

Section D (Vector Calculus)Q.No.9. (a) Discuss the physical interpretation of gradient of a scalar point function.

Also write a unit normal vector of a surface z = g(x, y). (2.5+0.5) Sol.: Geometrical interpretation:

1. is normal to the surface f (x, y, z) = c.

2. magnitude of is equal to the rate of change of f along this normal.

Consider the scalar point function , where .

Draw a surface f (x, y, z) = c through any point s.t. at each point on it, the function has the same value as at P. This type of surface is called a level surface of the function f through P. Examples: equipotential or isothermal surfaces are examples of level surfaces.

M P(R) O R Thus, if represents potential at the point , the equipotential surface =c is a level surface.

1. is normal to the surface f (x, y, z) = c.

Let be a point on a neighbouring level surface, where the function is .

Then .

Now, if lies on the same level surface as P, then

Now since .

This means thatis to every lying on this surface.

Thus is normal to the surface f (x, y, z) =c.

Now, if is unit vector normal to the surface f (x, y, z) =c, then we can write

.

is normal to the surface f (x, y, z) = c.

2. magnitude of is equal to the rate of change of f along this normal.

Let the perpendicular distance PM between the surfaces through P and is.

Then the rate of change of f normal to the surface through P.

Now

Hence, the magnitude of .

Thus the magnitude of is equal to the rate of change of f along this normal.Thus, grad (f) is a vector normal to the surface (f = constant) and has a magnitude equal to the rate of change of f along the normal.

2nd Part: A unit normal vector of a surface z = g(x, y) is

y-axisz-axisx-axis(2, 0, 0)(0, 2, 0)(0, 0, 4)ABCOQ.No.9. (b) Using the concept of surface integral, evaluate,

where and S is the triangular

surface lying in the first octant. (4)

Sol.: Since is a unit vector normal to the surface ,

i.e. .

.

.

Also .

Hence ,

where S is the surface and this surface lying in the first octant.

. Ans.

Q.No.10. (a) Show that may be written in the form

,where w(x, y) be a function which is continuous and has continuous first and second order partial derivatives in a domain of the xy-plane containing a region E of the type indicated in Greens theorem. Here s is the arc length of C. (4)Sol.: Let w(x, y) be a function which is continuous and has continuous first and second partial derivatives in a domain of the xy-plane containing a region E of the type indicated in Greens theorem.

We set and .

Then and are continuous in E.

Now since .

Using the expressions for and , we obtain

, the Laplacian of w. (i)

, where s is the arc length of C.The integrand of the last integral may be written as dot product of the vectors

.

Thus . (ii)

The vector is a unit normal vector to C, because the vector is a unit tangent vector to C, and .It follows that the expression on the RHS of (ii) is the derivative of w in the direction of the outward normal to C.

Denoting this directional derivative by . Then we obtain from Greens theorem the useful integral formula

. This completes the proof.

Q.No.10. (b) State divergence theorem and hence evaluate

where and S is the upper part of the sphere

above XOY plane. (1+3)Sol.: Applying Gauss Divergence theorem, which states that if F is a continuously differentiable vector function in the region E bounded by the closed surface S, then

,

where is the unit external normal vector at any point of S.

Here ,where S is the upper part of the sphere and S1 is the lower part of the sphere i.e. circle.

y-axisz-axisx-axisO SHere .

Put , we get

Q.No.12345

Ans.bdcba d e

Q.No. 6: 66 Q.No. 7: Q.No. 8:

Q.No. 9: Q.No. 10: True

Q.No.12345

Ans.bdcba d e

Q.No. 6: 66 Q.No. 7: Q.No. 8:

Q.No. 9: Q.No. 10: True

Q.No.12345

Ans.bdcba d e

Q.No. 6: 66 Q.No. 7: Q.No. 8:

Q.No. 9: Q.No. 10: True

Q.No.12345

Ans.bdcba d e

Q.No. 6: 66 Q.No. 7: Q.No. 8:

Q.No. 9: Q.No. 10: True

Q.No.12345

Ans.bdcba d e

Q.No. 6: 66 Q.No. 7: Q.No. 8:

Q.No. 9: Q.No. 10: True