If u = u(x, y) = sin(y + ax) - (y + ax)^2, th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $v = y + ax$. Then $u = \sin(v) - v^2$. First, we find the partial derivatives with respect to $x$: $u_x = \frac{\partial u}{\partial v} \cdot

Vedclass · Accepted Answer

$u_{xx} = a^2 u_{yy}$

Vedclass · Answer

(A) Let $v = y + ax$. Then $u = \sin(v) - v^2$. First, we find the partial derivatives with respect to $x$: $u_x = \frac{\partial u}{\partial v} \cdot \frac{\partial v}{\partial x} = (\cos(v) - 2v) \cdot a$. $u_{xx} = \frac{\partial}{\partial x} [a(\cos(v) - 2v)] = a(-\sin(v) - 2) \cdot \frac{\partial v}{\partial x} = a^2(-\sin(v) - 2)$. Next, we find the partial derivatives with respect to $y$: $u_y = \frac{\partial u}{\partial v} \cdot \frac{\partial v}{\partial y} = (\cos(v) - 2v) \cdot 1$. $u_{yy} = \frac{\partial}{\partial y} [\cos(v) - 2v] = (-\sin(v) - 2) \cdot \frac{\partial v}{\partial y} = -\sin(v) - 2$. Comparing $u_{xx}$ and $u_{yy}$, we see that $u_{xx} = a^2 u_{yy}$.

If $u = u(x, y) = \sin(y + ax) - (y + ax)^2$, then which of the following is true?

Similar Questions

If $u = \tan^{-1}\left(\frac{x^3 + y^3}{x - y}\right)$,then $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = $

The least value of $2x^{2} + y^{2} + 2xy + 2x - 3y + 8$ for real numbers $x$ and $y$ is:

If $u(x,y) = y \log x + x \log y$,then ${u_x}{u_y} - {u_x} \log x - {u_y} \log y + \log x \log y = $

If $u=f(r)$, where $r^2=x^2+y^2$, then $\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)$ is equal to

If $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$, then $\left. \frac{\partial f}{\partial x} \right|_{y = \frac{x}{4}}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module