Jacobians Questions and Answers

This set of Differential and Integral Calculus Multiple Choice Questions & Answers (MCQs) focuses on “Jacobians”.

1. The jacobian of p,q,r w.r.t x,y,z given p=x+y+z, q=y+z, r=z is ________
a) 0
b) 1
c) 2
d) -1
View Answer

Answer: b
Explanation: We have to find
\(J = \frac{∂(p,q,r)}{∂(x,y,z)} = \begin{vmatrix}
\frac{∂p}{∂x} & \frac{∂p}{∂y} &\frac{∂p}{∂z}\\
\frac{∂q}{∂x} &\frac{∂q}{∂y} &\frac{∂q}{∂z}\\
\frac{∂r}{∂x} &\frac{∂r}{∂y} &\frac{∂r}{∂z}\\
\end{vmatrix}\)
But p=x+y+z, q=y+z, r=z (taking partial derivative)
\(J=\begin{vmatrix}
1&1&1\\
0&1&1\\
0&0&1\\
\end{vmatrix}(\frac{∂p}{∂x}=1, \frac{∂p}{∂y}=1, \frac{∂p}{∂z}=1, \frac{∂q}{∂x}=0, \frac{∂q}{∂y}=1, \frac{∂q}{∂z}=1, \frac{∂r}{∂x}=0, \\
\frac{∂r}{∂y}=0, \frac{∂r}{∂z}=1)\)
On expanding we get
J = 1(1 – 0) = 1
Thus j = 1.

2. Given \(u=\frac{yz}{x}, v=\frac{zx}{y}, w=\frac{xy}{z}\) then the value of \(\frac{∂(u,v,w)}{∂(x,y,z)}\) is ________
a) 4
b) -4
c) 0
d) 1
View Answer

Answer: a
Explanation: By Data \(u=\frac{yz}{x}, v=\frac{zx}{y}, w=\frac{xy}{z}\)

\(J = \frac{∂(u,v,w)}{∂(x,y,z)} = \begin{vmatrix}
\frac{∂u}{∂x} & \frac{∂u}{∂y} &\frac{∂u}{∂z}\\
\frac{∂v}{∂x} &\frac{∂v}{∂y} &\frac{∂v}{∂z}\\
\frac{∂w}{∂x} &\frac{∂w}{∂y} &\frac{∂w}{∂z}\\
\end{vmatrix} = \begin{vmatrix}
\frac{-yz}{x^2} & \frac{z}{x} &\frac{y}{x}\\
\frac{z}{y} &\frac{-zx}{y^2} &\frac{x}{y}\\
\frac{y}{z} &\frac{x}{z} &\frac{-xy}{z^2}\\
\end{vmatrix}\)
\(=\frac{-yz}{x^2}\Big\{(\frac{-zx}{y^2})(\frac{-xy}{y^2}) – (\frac{x}{z})(\frac{x}{y})\Big\} – (\frac{z}{x})\Big\{(\frac{z}{y})(\frac{-xy}{z^2})-(\frac{y}{z})(\frac{x}{y})\Big\} \\
+ \frac{y}{x}\Big\{(\frac{z}{y})(\frac{x}{z})-(\frac{y}{z})(\frac{-zx}{y^2})\Big\}\)
\(=\frac{-yz}{x^2} \Big\{\frac{x^2}{yz} – \frac{x^2}{yz}\Big\} – \frac{z}{x}\Big\{\frac{-x}{z} – \frac{x}{z}\Big\} + \frac{y}{x}\Big\{\frac{x}{y} + \frac{x}{y}\Big\} = 0 + 1 + 1 + 1 = 4\)
Therefore \(\frac{∂(u,v,w)}{∂(x,y,z)} = 4.\)

3. If u=x+3y²-z³, v=4x² yz, w=2z²-xy then \(\frac{∂(u,v,w)}{∂(x,y,z)} \) at (1,1,1).
a) -184
b) -90
c) 20
d) 40
View Answer

Answer: a
Explanation: Given that u=x+3y²-z³, v=4x² yz, w=2z²-xy
\(\frac{∂(u,v,w)}{∂(x,y,z)} = \begin{vmatrix}
\frac{∂u}{∂x} & \frac{∂u}{∂y} &\frac{∂u}{∂z}\\
\frac{∂v}{∂x} &\frac{∂v}{∂y} &\frac{∂v}{∂z}\\
\frac{∂w}{∂x} &\frac{∂w}{∂y} &\frac{∂w}{∂z}\\
\end{vmatrix} = \begin{vmatrix}
1 & 6y & -3z^2\\
8xyz &4x^2 z &4x^2 y\\
-y &-x &4z\\
\end{vmatrix}\)
\(\frac{∂(u,v,w)}{∂(x,y,z)} \,at\, (1,1,1) = \begin{vmatrix}
1 & 6 & -3\\
8 &4&4\\
-1 &-1 &4\\
\end{vmatrix}\)
=1(16+4) – 6(32+4) – 3(-8+4) = -184.

4. If x=rcos⁡θ, y=rsin⁡θ then the value of \(\frac{∂(x,y)}{∂(r,θ)}\) is ________
a) 1
b) 0
c) r
d) \(\frac{1}{r}\)
View Answer

Answer: c
Explanation: Wkt, \(\frac{∂(x,y)}{∂(r,θ)} = \begin{vmatrix}
\frac{∂x}{∂r} & \frac{∂x}{∂θ}\\
\frac{∂y}{∂r} & \frac{∂y}{∂θ}\\
\end{vmatrix} \,but\, x=rcos⁡θ, y=rsin⁡θ\)
Thus \( \begin{vmatrix}
cos⁡θ & -rsin⁡θ\\
sin⁡θ & rcos⁡θ\\
\end{vmatrix} = rcos^2 θ + rsin^2 θ = r.\)

5. The application of Jacobians is significant in the evaluation of double integral of the form ∬f(x,y) dx dy and triple integral of the form ∭f(x,y,z)dx dy dz by transformation from one system of coordinate to the other.
a) True
b) False
View Answer

Answer: a
Explanation: The principle of evaluation is analogous with the evaluation of ∫f(x)dx by taking a suitable substitution.

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6. If u+v=e^x cos⁡y and u-v=e^x sin y the value of \(J(\frac{u,v}{x,y})\) is ________
a) e^2x
b) \(\frac{e^2x}{2} \)
c) \(\frac{-e^2x}{2} \)
d) 0
View Answer

Answer: c
Explanation: wkt \(J(\frac{u,v}{x,y}) = \frac{∂(u,v)}{∂(x,y)} = \begin{vmatrix}
\frac{∂u}{∂x} & \frac{∂u}{∂y}\\
\frac{∂v}{∂x} & \frac{∂v}{∂y}\\
\end{vmatrix}\)
But we have to find u and v first using given equation
u+v=e^x cos ⁡y………(1)
u-v=e^x sin⁡ y……..(2)
solving (1)&(2)
we get u=\(\frac{e^x}{2} \)(cos⁡y+sin⁡y)
v=\(\frac{e^x}{2} \)(cos⁡y-sin⁡y)
\(\frac{∂u}{∂x} = \frac{e^x}{2}(cos⁡y+sin⁡y) \)
\(\frac{∂v}{∂x} = \frac{e^x}{2}(cos⁡y-sin⁡y) \)
\(\frac{∂u}{∂y} = \frac{e^x}{2}(cos⁡y-sin⁡y) \)
\(\frac{∂v}{∂y} = \frac{e^x}{2}(-cos⁡y-sin⁡y) \)
\(J(\frac{u,v}{x,y}) = \begin{vmatrix}
\frac{e^x}{2}(cos⁡y+sin⁡y) & \frac{e^x}{2}(cos⁡y-sin⁡y)\\
\frac{e^x}{2}(cos⁡y-sin⁡y) & \frac{e^x}{2}(-cos⁡y-sin⁡y)\\
\end{vmatrix}\)
\((\frac{e^x}{2})(\frac{e^x}{2})\){-(cos⁡y+sin⁡y)² -(cos⁡y-sin⁡y)²}…….(expanding & solving by taking cos² y+sin² y=1)
=\(\frac{-e^2x}{2} \).

7. Which among the following is the definition of Jacobian of u and v w.r.t x and y?
a) \(J(\frac{x,y}{u,v})\)
b) \(J(\frac{u,v}{x,y})\)
c) \(\frac{∂(x,y)}{∂(u,v)} \)
d) \(\frac{∂(u,x)}{∂(v,y)} \)
View Answer

Answer: b
Explanation: \(J(\frac{u,v}{x,y}) = \frac{∂(u,v)}{∂(x,y)} = \begin{vmatrix}
\frac{∂u}{∂x} & \frac{∂u}{∂y}\\
\frac{∂v}{∂x} & \frac{∂v}{∂y}\\
\end{vmatrix}\)

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