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
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 logx-k = 0
b) x2+2y2–\(\frac{a^2}{2}\) logx-k = 0
c) x2+y2-2a2 logx-k = 0
d) 2x2-y2–\(\frac{a^2}{3}\) logx-k = 0
View Answer
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 logx – 2c=0 or x2+y2-2a2 logx-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
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
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, \( \frac{1}{r} \frac{dr}{dθ} = \frac{cosθ}{1+sinθ}\) 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 \(-r\frac{dθ}{dr} = \frac{cosθ}{1+sinθ}\) …..separating the variables and integrating it
\(\int \frac{dr}{r} + \int \frac{1+sinθ}{cosθ} \,dθ = c \rightarrow log r + \int secθ \,dθ + \int tanθ \, dθ = c\)
log r + log(secθ + tanθ) + log(secθ) = c = log k
\(\rightarrow log(r(secθ + tanθ) (secθ)) = log k \rightarrow r\left(\frac{1}{cosθ} + \frac{sinθ}{cosθ}\right) \frac{1}{cosθ} = k\)
\(\frac{r(1+sinθ)}{cos^2 θ} = \frac{r(1+sinθ)}{1-sin^2 θ} \rightarrow r = k(1-sin θ)\) is the required orthogonal trajectory.
5. Which among the following is true for the curve rn = a sinnθ?
a) Given family of curve is Self orthogonal
b) Orthogonal trajectory is rn=k cosnθ where k is an constant
c) Orthogonal trajectory is rn=k cosecnθ where k is an constant
d) Orthogonal trajectory is rn=k sinnθ where k is an constant
View Answer
Explanation: Consider rn = a sinnθ..(1) –> n log r = log a + log(sinnθ) …differentiating w.r.t θ
\( \frac{n}{r} \frac{dr}{dθ} = \frac{n cosnθ}{sinnθ} \,or\, \frac{1}{r} \frac{dr}{dθ} = \frac{cosnθ}{sinnθ} = cot \,nθ\)
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 \(\int \frac{dr}{r} \int tan\,nθ \,dθ = c \)
\( \rightarrow log \,r + \frac{log(secnθ)}{n} = c\) or n log r + log(sec nθ) = nc = logk
log(rn secnθ) = log k –> rn sec nθ=k or rn=k cosnθ ..(2) is the required orthogonal trajectory since (1)&(2) are not same the given family of curve is not self orthogonal.
Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.
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