Question

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

This question was addressed to me in an internship interview.

My query is from Jacobians topic in portion Partial Differentiation of Engineering Mathematics

Answer 1

Correct choice is (c) \(\frac{-e^2x}{2} \)

To elaborate: 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 cos^2 y+sin^2 y=1)

=\(\frac{-e^2x}{2} \).