# Differential and Integral Calculus Questions and Answers – Jacobians

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

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

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

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) $$\frac{1}{r}$$

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

Explanation: The principle of evaluation is analogous with the evaluation of ∫f(x)dx by taking a suitable substitution.
Note: Join free Sanfoundry classes at Telegram or Youtube

6. If u+v=ex cos⁡y and u-v=ex sin y the value of $$J(\frac{u,v}{x,y})$$ is ________
a) e2x
b) $$\frac{e^2x}{2}$$
c) $$\frac{-e^2x}{2}$$
d) 0

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=ex cos ⁡y………(1)
u-v=ex 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) $$\frac{∂(u,x)}{∂(v,y)}$$

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

Sanfoundry Global Education & Learning Series – Differential and Integral Calculus.

To practice all areas of Differential and Integral Calculus, here is complete set of 1000+ Multiple Choice Questions and Answers.