JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech I Year MATHEMATICS-II Previous Question Paper (September/October - 2021 - R18)

B.Tech | I Year | II Semester | R18 | September/October - 2021

Answer PaperPaper Code: 152AAMaximum Marks: 75Time: 3 Hours

Answer any five questions. All questions carry equal marks.

Main

115 marks

Solve the differential equation

dy/dx = (y + x sin x) / cos x

7 marks

Answer

The given differential equation is a first-order linear differential equation of the form

dy/dx + P(x)y = Q(x)

. Rearranging the given equation:

dy/dx - (1/cos x)y = (x sin x)/cos x

This simplifies to:

dy/dx - (sec x)y = x tan x

Here, P(x) = -sec x and Q(x) = x tan x. The integrating factor (IF) is:

IF = e^{∫ -sec x dx} = e^{-ln|sec x + tan x|} = 1/(sec x + tan x) = sec x - tan x

The general solution is given by y(IF) = ∫ Q(x)(IF) dx + C:

y(sec x - tan x) = ∫ (x tan x)(sec x - tan x) dx + C

y(sec x - tan x) = ∫ (x sec x tan x - x tan² x) dx + C

y(sec x - tan x) = ∫ (x sec x tan x - x(sec² x - 1)) dx + C

y(sec x - tan x) = ∫ (x sec x tan x - x sec² x + x) dx + C

Using integration by parts for the first two terms, the final solution is:

y(sec x - tan x) = x sec x - ln|sec x + tan x| - (x tan x - ln|sec x|) + x²/2 + C

If the air is maintained at 15⁰C and the temperature of the body drops from 70⁰C to 40⁰C in 10 minutes. What will be its temperature after 30 minutes?

8 marks

Answer

According to Newton's Law of Cooling, the rate of change of temperature is proportional to the difference between the body temperature T and the ambient temperature T_s:

dT/dt = -k(T - T_s)

Given T_s = 15°C, the solution is T(t) = 15 + Ce^-kt. At t = 0, T = 70:

70 = 15 + C ⇒ C = 55

At t = 10, T = 40:

40 = 15 + 55e^-10k ⇒ 25/55 = e^-10k ⇒ e^-10k = 5/11

To find the temperature at t = 30 minutes:

T(30) = 15 + 55(e^-10k)³

T(30) = 15 + 55(5/11)³

T(30) = 15 + 55(125/1331) = 15 + 6875/1331 ≈ 15 + 5.166 = 20.166°C

215 marks

Solve the differential equation

(1 + e^x) cos y dx + e^x sin y dy = 0

7 marks

Answer

The temperature of cup of coffee is 92⁰C. When freshly poured the room temperature being 24⁰C. In one minute it was cooled to 80⁰C. How long a period must elapse, before the temperature of the cup becomes 65⁰C?

8 marks

Answer

315 marks

Solve the differential equation

(D³ - 3D² + 3D - 1)y = Sin x + x³

7 marks

Answer

Solve by method of variation of parameters

(d²y/dx²) + y = cos x

8 marks

Answer

415 marks

Evaluate

∫∫ (x + y) dx dy

in the positive quadrant for which

x² + y² ≤ 1

8 marks

Answer

Evaluate

∫₀¹ ∫₀^1-x xy dx dy

7 marks

Answer

515 marks

By changing the order of integration, evaluate

∫₀¹ ∫₀^x xy dx dy

7 marks

Answer

Evaluate

∫₀¹ ∫₀^1-z ∫₀^1-y-z xyz dx dy dz

, where V is the region of space bounded by x = 0, x = 1, y = 0, y = 2, z = 0, z = 3

8 marks

Answer

615 marks

Find the angle between the surface x log z = y² - 1 and x² + y² + z = 2 at (1, 1, 1).

7 marks

Answer

Find work done in moving a particle in the force field F = (2x - y + z)i + (x + y - z²)j + (3x - 2y + 4z)k along the curve x = 2t², y = t, z = 4t² - 1 from t = 0 to 1.

8 marks

Answer

715 marks

Find the directional derivative of f(x,y,z) = zy³ + xz³ at the point (1,-3,4) in the direction of the vector 2i + j - 3k.

7 marks

Answer

Show that the vector F = (3x² + 6y)i - (14yz)j + 20xz²k is irrotational and find scalar potential.

8 marks

Answer

Verify Stokes theorem for F = (x² + y²)i - 2xyj taken around the rectangle bounded by the lines x = ±a, y = 0, y = b.

15 marks

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech I Year MATHEMATICS-II Previous Question Paper (September/October - 2021 - R18)

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