Ordinary Differential Equations Questions and Answers – Orthogonal Trajectories

This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “Orthogonal Trajectories”.

1. Find the orthogonal trajectories of the family of parabolas y2=4ax.
a) 2x2+y2=k
b) 2y2+x2=k
c) x2-2y2=k
d) 2x2-y2=k
View Answer

Answer: a
Explanation: Consider \(\frac{y^2}{x} = 4a\)……..differentiating w.r.t x we get \(\frac{x.2y \frac{dy}{dx} – y^2.1}{x^2} = 0\)
\(2xy \frac{dy}{dx} -y = 0\)…..DE of the family of curve y2=4ax replacing \(\frac{dy}{dx}\) by \(-\frac{dx}{dy}\)
since \(\frac{dy}{dx}\) is the slope of the tangent to the curve –> \(-\frac{dx}{dy}\) is the slope of orthogonal line we get \(2x(-\frac{dx}{dy}) – y = 0 \, or\, 2x \,dx + y \,dy = 0 \rightarrow \int2x \,dx + \int y \,dy = c \rightarrow x^2 + \frac{y^2}{2} = c.\) or 2x2+y2=k is the required orthogonal trajectory.

2. Find the orthogonal trajectories of the family of curves \(\frac{x^2}{a^2} + \frac{y^2}{b^2+k} = 1\) where k is the parameter.
a) x2-y2-3a2 log⁡x-k = 0
b) x2+2y2–\(\frac{a^2}{2}\) log⁡x-k = 0
c) x2+y2-2a2 log⁡x-k = 0
d) 2x2-y2–\(\frac{a^2}{3}\) log⁡x-k = 0
View Answer

Answer: c
Explanation: \(\frac{x^2}{a^2} + \frac{y^2}{b^2+k} = 1\)…….(1) differentiating w.r.t x we have \(\frac{2x}{a^2} + \frac{2yy’}{b^2+k} = 0 \,where\, y’ = \frac{dy}{dx}\)
i.e \(\frac{x^2}{a^2} = \frac{-yy’}{b^2+k} …..(2) \,from\, (1)\, \frac{x^2}{a^2} – 1 = \frac{-y^2}{b^2+k} \rightarrow \frac{x^2-a^2}{a^2} = \frac{-y^2}{b^2+k}\)…..(3) divide (2)&(3)
we get \(\frac{x}{x^2-a^2} = \frac{yy’}{y^2} \rightarrow \frac{x}{x^2-a^2} = \frac{y’}{y}\)
now \(y’=\frac{dy}{dx}\) is replaced by \(-\frac{dx}{dy} \,i.e\, \frac{x}{x^2-a^2} = \frac{1}{y}(-\frac{dx}{dy})\) separating the variable we get \(y \,dy = -\frac{(x^2-a^2)}{x} dx\) integrating this equation
\(\int y \,dy = \int-x \,dx + \int a^2 \frac{1}{x} \,dx + c \rightarrow \frac{y^2}{2} = \frac{-x^2}{2} + a^2 log⁡ \,x + c\)
–> x2+y2-2a2 log⁡x – 2c=0 or x2+y2-2a2 log⁡x-k=0 where k=2c is the required orthogonal trajectory.

3. The Orthogonal DE for family of parabola y2=4a(x+a) is same as _______(where DE stands for Differential equation)
a) DE of parabola y2=4a(x+a)
b) DE of parabola y2=4ax
c) DE of parabola x2=4ay
d) DE of parabola x2=4a(y+a)
View Answer

Answer: a
Explanation: y2=4a(x+a)……….differentiating w.r.t x we get \(2y \frac{dy}{dx} = 4a \rightarrow a =\frac{yy’}{2}\)
substituting the ‘a’ in given equation i.e : \(y^2 = 2yy’(x + \frac{yy’}{2}), y=2xy’+yy’^2\)…(1)
replacing y’ by \(\frac{-1}{y’} \,i.e\, y=2x\frac{-1}{y’} + y(\frac{-1}{y’})^2\) or on solving yy’2+2xy’=y…(2)
comparing (2) & (1) we observe that DE of the orthogonal family is same as the DE of the given family of curves thus the family of parabola is self orthogonal.

4. Find the orthogonal trajectories of the family r=a(1+sin θ).
a) r=k(sin θ)
b) r2=k(cos θ)2
c) r=k(1-cos θ)
d) r=k(1-sin θ)
View Answer

Answer: d
Explanation: r=a(1+sin θ) taking log we get log r=log a+log(1+sin θ)…differentiating
w.r.t θ we have, \( \frac{1}{r} \frac{dr}{dθ} = \frac{cos⁡θ}{1+sin⁡θ}\) since given equation is polar slope of tangent is given by \( \frac{1}{r} \frac{dr}{dθ}\) and perpendicular line has a slope of \(-r\frac{dθ}{dr} \, replacing\, \frac{1}{r} \frac{dr}{dθ} \,by\, -r\frac{dθ}{dr}\)
we have \(-r\frac{dθ}{dr} = \frac{cos⁡θ}{1+sin⁡θ}\) …..separating the variables and integrating it
\(\int \frac{dr}{r} + \int \frac{1+sin⁡θ}{cos⁡θ} \,dθ = c \rightarrow log r + \int sec⁡θ \,dθ + \int tan⁡θ \, dθ = c\)
log r + log(sec⁡θ + tan⁡θ) + log(sec⁡θ) = c = log k
\(\rightarrow log(r(sec⁡θ + tan⁡θ) (sec⁡θ)) = log k \rightarrow r\left(\frac{1}{cos⁡θ} + \frac{sin⁡θ}{cos⁡θ}\right) \frac{1}{cos⁡θ} = k\)
\(\frac{r(1+sin⁡θ)}{cos^2 θ} = \frac{r(1+sin⁡θ)}{1-sin^2 θ} \rightarrow r = k(1-sin θ)\) is the required orthogonal trajectory.

5. Which among the following is true for the curve rn = a sin⁡nθ?
a) Given family of curve is Self orthogonal
b) Orthogonal trajectory is rn=k cos⁡nθ where k is an constant
c) Orthogonal trajectory is rn=k cosec⁡nθ where k is an constant
d) Orthogonal trajectory is rn=k sin⁡nθ where k is an constant
View Answer

Answer: b
Explanation: Consider rn = a sin⁡nθ..(1) –> n log r = log a + log(sin⁡nθ) …differentiating w.r.t θ
\( \frac{n}{r} \frac{dr}{dθ} = \frac{n cos⁡nθ}{sin⁡nθ} \,or\, \frac{1}{r} \frac{dr}{dθ} = \frac{cos⁡nθ}{sin⁡nθ} = cot \,nθ\)
replacing \(\frac{1}{r} \frac{dr}{dθ} \,by\, -r\frac{dθ}{dr} \,we\,\, get\, -r\frac{dθ}{dr} = cot \,nθ\)
separating the variables and integrating \(\int \frac{dr}{r} \int tan⁡\,nθ \,dθ = c \)
\( \rightarrow log \,r + \frac{log⁡(sec⁡nθ)}{n} = c\) or n log r + log⁡(sec nθ) = nc = log⁡k
log(rn sec⁡nθ) = log k –> rn sec⁡ nθ=k or rn=k cos⁡nθ ..(2) is the required orthogonal trajectory since (1)&(2) are not same the given family of curve is not self orthogonal.
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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