Paper I — Q6
(a) If F(s) and G(s) are Laplace transforms of f(t) and g(t) respectively, then prove that L∫₀^t f(x) g(t-x) dx = F(s)…
If F(s) and G(s) are Laplace transforms of f(t) and g(t) respectively, then prove that
L∫₀^t f(x) g(t-x) dx = F(s) G(s).
Using this result, solve the equation
y(t) = t + ∫₀^t y(x) sin(t-x) dx. 15 marks
One end of an elastic string, having natural length a, is fixed at some point O and a heavy particle is attached to the other end of the string. The string is drawn vertically downward till it is four times its natural length at the point C and then released. If the modulus of elasticity of the string is equal to the weight of the particle, then show that the particle will return to the same point C in the time
√(a/g)(2√3 + (4π)/3). 15 marks
Find the absolute value of the directional derivative of φ(x, y, z) = x^2y^2z² at the point (1, 1, -1) in the direction of the tangent to the curve x = e^t, y = 2sin t + 1, z = t - cos t, at t = 0. 10 marks
If ∇ · overrightarrowE=0, ∇ · overrightarrowH=0, ∇ × overrightarrowE=-(∂ overrightarrowH)/(∂ t) and ∇ × overrightarrowH=(∂ overrightarrowE)/(∂ t),
then show that ∇² overrightarrowH=(∂² overrightarrowH)/(∂ t²) and ∇² overrightarrowE=(∂² overrightarrowE)/(∂ t²). 10 marks
हिंदी में प्रश्न पढ़ें
यदि f(t) और g(t) के लाप्लास रूपान्तर क्रमशः F(s) और G(s) हैं, तो सिद्ध कीजिए कि
L∫₀^t f(x) g(t-x) dx = F(s) G(s)
है। इस परिणाम का प्रयोग करते हुए, समीकरण
y(t) = t + ∫₀^t y(x) sin(t-x) dx
को हल कीजिए। 15 अंक
एक प्रत्यास्थ डोरी, जिसकी प्राकृतिक लंबाई a है, का एक छोर किसि बिंदु O पर स्थिर है और डोरी के दूसरे छोर पर एक भारी कण जुड़ा हुआ है। डोरी को उर्ध्वाधर नीचे की ओर बिंदु C तक तब तक खींचा जाता है जब तक वह अपनी प्राकृतिक लंबाई से चार गुना न हो जाए तथा फिर छोड़ दिया जाता है। यदि डोरी का प्रत्यास्थता गुणांक कण के भार के बराबर है, तो दर्शाइए कि कण
√(a/g)(2√3 + (4π)/3)
समय में उसी बिंदु C पर वापस आ जाएगा। 15 अंक
φ(x, y, z) = x^2y^2z² का बिंदु (1, 1, -1) पर, वक्र x = e^t, y = 2sin t + 1, z = t - cos t, के बिंदु t = 0 पर स्पर्श-रेखा की दिशा में दिक्-अवकलज का निरपेक्ष मान ज्ञात कीजिए। 10 अंक
यदि ∇ · overrightarrowE=0, ∇ · overrightarrowH=0, ∇ × overrightarrowE=-(∂ overrightarrowH)/(∂ t) और ∇ × overrightarrowH=(∂ overrightarrowE)/(∂ t) है,
तो दर्शाइए कि ∇² overrightarrowH=(∂² overrightarrowH)/(∂ t²) और ∇² overrightarrowE=(∂² overrightarrowE)/(∂ t²) है। 10 अंक
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a)
Let F(s)=∫ from 0 to ∞ e^(−st)f(t)dt and G(s)=∫ from 0 to ∞ e^(−st)g(t)dt, assuming f and g are piecewise continuous and of exponential order so the transforms exist. By definition,
L{∫₀ᵗ f(x)g(t−x)dx}=∫ from 0 to ∞ e^(−st)∫₀ᵗ f(x)g(t−x)dx dt.
Change the order of integration. Here x runs from 0 to ∞ and, for fixed x, t runs from x to ∞. Put u=t−x, so dt=du and t=x+u. Then
=∫₀∞ f(x)∫₀∞ e^(−s(x+u))g(u)du dx =(∫₀∞ e^(−sx)f(x)dx)(∫₀∞ e^(−su)g(u)du) =F(s)G(s).
This proves the convolution theorem.
Now take Laplace transform of
y(t)=t+∫₀ᵗ y(x)sin(t−x)dx.
The integral is y(t) convolved with sin t. Hence
Y(s)=1/s²+Y(s)L{sin t}=1/s²+Y(s)/(s²+1).
Therefore
Y(s)(1−1/(s²+1))=1/s², Y(s)= (s²+1)/s⁴=1/s²+1/s⁴.
Using L⁻¹{1/s²}=t and L⁻¹{1/s⁴}=t³/6,
y(t)=t+t³/6.
(b)
Let m be the mass of the particle, so its weight is mg. The modulus of elasticity is given equal to the weight, so λ=mg. Let y be the distance of the particle below O. For y≥a, the tension is
T=λ(y−a)/a=mg(y−a)/a.
The equation of motion downward is
mÿ=mg−T=mg−mg(y−a)/a=(mg/a)(2a−y).
Put X=y−2a. Then
Ẍ=−(g/a)X.
Thus the motion is simple harmonic about y=2a with angular frequency ω=√(g/a). At C, y=4a, so X=2a. The particle is released from rest there. Therefore
X=2a cos ωt.
The string becomes slack when y=a, i.e. X=−a. Then
2a cos ωt=−a, so cos ωt=−1/2.
The first such time is
t₁=2π/(3ω)=(2π/3)√(a/g).
At this instant,
Ẋ=−2aω sin(2π/3)=−√(3ag),
so the particle is moving upward with speed √(3ag).
During the slack phase, y<a and T=0, so only gravity acts. The time to go up to the highest point and return to y=a is
t₂=2(√(3ag)/g)=2√(3a/g).
The highest point is y=a−3a/2=−a/2, so the string remains slack throughout this phase.
When the particle returns to y=a, its downward speed is again √(3ag). The string becomes taut and SHM resumes. At X=−a with Ẋ=+√(3ag), the amplitude is again 2a and the initial phase is 4π/3. Hence the time to reach C, where X=2a, is
t₃=2π/(3ω)=(2π/3)√(a/g).
Total time to first return to C is
t₁+t₂+t₃=√(a/g)(2√3+4π/3).
Hence shown.
(c)(i)
Given φ=x²y²z². At P(1,1,−1),
∇φ=(∂φ/∂x,∂φ/∂y,∂φ/∂z) =(2xy²z²,2x²yz²,2x²y²z) =(2,2,−2).
The curve is r(t)=(eᵗ,2sin t+1,t−cos t). Its tangent vector is
r′(t)=(eᵗ,2cos t,1+sin t).
At t=0,
r′(0)=(1,2,1).
The unit tangent is
u=(1,2,1)/√6.
Thus the directional derivative is
∇φ·u=(2+4−2)/√6=4/√6=2√6/3.
Therefore, its absolute value is
2√6/3.
(c)(ii)
Use the vector identity
∇×(∇×A)=∇(∇·A)−∇²A.
For E,
∇×(∇×E)=∇(∇·E)−∇²E.
But
∇×(∇×E)=∇×(−∂H/∂t) =−∂/∂t(∇×H) =−∂/∂t(∂E/∂t) =−∂²E/∂t².
Since ∇·E=0, the left side is −∇²E. Hence
−∇²E=−∂²E/∂t²,
so
∇²E=∂²E/∂t².
For H,
∇×(∇×H)=∇(∇·H)−∇²H.
Also,
∇×(∇×H)=∇×(∂E/∂t) =∂/∂t(∇×E) =∂/∂t(−∂H/∂t) =−∂²H/∂t².
Since ∇·H=0, the left side is −∇²H. Therefore,
−∇²H=−∂²H/∂t²,
so
∇²H=∂²H/∂t².
How this answer will be evaluated
What earns full marks: All parts solved with correct methods and final answers.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Prove the convolution theorem and solve the integral equation. 15 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Prove convolution theorem via double integral and change of order (u = t−x)
- Apply to y(t) = t + (y ∗ sin)(t)
- Y(s) = 1/s² + Y(s)/(s²+1) ⇒ Y(s) = (s²+1)/s⁴
- Invert: y(t) = t + t³/6
Loses marks
- Skipping the proof of the convolution theorem
- Incorrect algebraic simplification of Y(s)
- Failing to invert the Laplace transform correctly
Earns more
- Recognises the integral as the convolution y ∗ sin t, with L{t} = 1/s² and L{sin t} = 1/(s²+1)
- Simplifies Y(s) to (s²+1)/s⁴
- Splits Y(s) as 1/s² + 1/s⁴ before inverting
- Final answer is y(t) = t + t³/6
Extra mark
- Verifies the solution by substituting back into the integral
- (b) Show the particle returns to point C in the specified time. 15 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- λ = mg; mÿ = (mg/a)(2a − y); SHM about y = 2a with ω = √(g/a)
- First phase to slack point y = a: t₁ = 2π/(3ω); speed at slack √(3ag)
- Slack free flight t₂ = 2√(3a/g); apex y = −a/2 (string stays slack)
- Third SHM phase t₃ = 2π/(3ω); total = √(a/g)(2√3 + 4π/3)
Loses marks
- Incorrect setup of the equation of motion
- Failing to account for the equilibrium position
- Incorrect calculation of the time periods
Earns more
- Correctly identifies the modulus of elasticity condition
- Derives the SHM equation Ẍ = −(g/a)X about the equilibrium y = 2a
- Time in the slack (free-flight) phase = 2√3·√(a/g)
- Time in the two elastic phases together = (4π/3)√(a/g)
Extra mark
- Draws a neat diagram of the elastic string and particle
- (c)(i) Find the absolute value of the directional derivative. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- ∇φ at (1,1,−1) = (2,2,−2)
- Tangent r′(0) = (1,2,1); unit tangent (1,2,1)/√6
- Directional derivative = (2+4−2)/√6 = 4/√6
- Absolute value 2√6/3 ≈ 1.633
Loses marks
- Incorrect calculation of the gradient
- Failing to normalize the tangent vector
- Incorrect dot product calculation
Earns more
- Correctly calculates the gradient as (2, 2, −2)
- Finds the tangent vector as (1, 2, 1)
- Normalizes the tangent vector to (1/√6, 2/√6, 1/√6)
- Final answer is 4/√6 (or 2√6/3)
Extra mark
- Verifies the result using an alternative method
- (c)(ii) Show the wave equations for E and H. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Use ∇×(∇×A) = ∇(∇·A) − ∇²A
- For E: −∇²E = −∂²E/∂t² ⇒ ∇²E = ∂²E/∂t²
- For H: −∇²H = −∂²H/∂t² ⇒ ∇²H = ∂²H/∂t²
- Both wave equations derived
Loses marks
- Incorrect application of the vector identity
- Failing to substitute the given Maxwell's equations
- Incorrect simplification of the expressions
Earns more
- Correctly applies the vector identity
- Substitutes ∇·E = 0 and ∇·H = 0
- Simplifies the expressions correctly
- Final results are ∇²E = ∂²E/∂t² and ∇²H = ∂²H/∂t²
Extra mark
- Mentions the physical significance of the wave equations
What "Prove" is asking you to do
Establish that the statement holds for every case it claims, not for one representative case. The argument must be closed: each line follows from a definition, a hypothesis, or a named theorem you are entitled to use.
Structure that answers it
Given and to prove, restated → theorem or construction to be used, named → the argument line by line → conclusion stated as proved
Where marks are lost
Testing one example, which illustrates but proves nothing. On an if and only if claim, proving one direction and stopping forfeits that half outright, and degenerate cases — zero, the empty set, the equality case — have to be disposed of rather than assumed away.
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Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.
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