Chapter 15, Problem 1RE

(a)

To determine

Whether the statement “The level curves of $g\left(x,y\right)=e^{x+y}$ are lines” is true or false.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

The given function is, $g\left(x,y\right)=e^{x+y}$.

Let $e^{x+y}=k$.

Take log on both sides.

$\begin{array}{l}{e}^{x+y}=k\\ \ln \left(e^{x+y}\right)=\ln \left(k\right)\\ x+y=\ln k\left(\because {\log }_a\left(a^b\right)=b\right)\end{array}$

Here, $x+y=\ln k$ is the equation of a line.

Therefore, the statement is true.

(b)

To determine

Whether the equation $z^2=2x^2-6y^2$ satisfies z as a single function of x and y.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Given:

The equation is $z^2=2x^2-6y^2$.

Calculation:

The given equation is $z^2=2x^2-6y^2$.

When $2x^2-6y^2>0$, the equation $z^2=2x^2-6y^2$ becomes,

$\begin{array}{l}{z}^2=2x^2-6y^2\\ z=\pm \sqrt{2x^2-6y^2}\\ z=\sqrt{2x^2-6y^2},z=-\sqrt{2x^2-6y^2}\end{array}$

The functions are $z\left(x,y\right)=\sqrt{2x^2-6y^2}$ and $z\left(x,y\right)=-\sqrt{2x^2-6y^2}$ in terms of x and y.

Therefore, the statement is false.

(c)

To determine

Whether the function f satisfies the derivative $f_{xxy}=f_{yyx}$.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Let the function f has a continuous partial derivatives of all orders.

Then prove that $f_{xxy}=f_{yyx}$.

For example, assume $f\left(x,y\right)=x^{2}y$.

Obtain the value of $f_{xxy}$.

Take partial derivative of the function f with respect to x and obtain $f_x$.

$\begin{array}{c}{f}_x=\frac{\partial }{\partial x}\left(x^{2}y\right)\\ =y\frac{\partial }{\partial x}\left(x^2\right)\\ =y\left(2x\right)\\ =2xy\end{array}$

Thus, $f_x=2xy$.                                                                                                        (1)

Take partial derivative of the equation (1) with respect to x and obtain $f_{xx}$.

$\begin{array}{c}{f}_{xx}=\frac{\partial }{\partial x}\left(2xy\right)\\ =2y\frac{\partial }{\partial x}\left(x\right)\\ =2y\left(1\right)\\ =2y\end{array}$

Hence, $f_{xx}=2y$.                                                                                                     (2)

Again, take partial derivative for the equation (2) with respect to y and obtain $f_{xxy}$.

$\begin{array}{c}{f}_{xxy}=\frac{\partial }{\partial y}\left(2y\right)\\ =2\frac{\partial }{\partial y}\left(y\right)\\ =2\left(1\right)\\ =2\end{array}$

Therefore, $f_{xxy}=2$.

Obtain the value of $f_{yyx}$.

Take partial derivative of the function f with respect to y and obtain $f_y$.

$\begin{array}{c}{f}_y=\frac{\partial }{\partial y}\left(x^{2}y\right)\\ =x^2\frac{\partial }{\partial y}\left(y\right)\\ =x^2\left(1\right)\\ =x^2\end{array}$

Thus, $f_y=x^2$.                                                                                                          (3)

Take partial derivative of the equation (1) with respect to y and obtain $f_{yy}$.

$\begin{array}{c}{f}_{yy}=\frac{\partial }{\partial y}\left(x^2\right)\\ =x^2\frac{\partial }{\partial y}\left(1\right)\\ =x^2\left(0\right)\\ =0\end{array}$

Hence, $f_{yy}=0$.                                                                                                        (4)

Again, take partial derivative for the equation (2) with respect to x and obtain $f_{yyx}$.

$\begin{array}{c}{f}_{yyx}=\frac{\partial }{\partial x}\left(0\right)\\ =0\end{array}$

Therefore, $f_{yyx}=0$.

From above, it is concluded that $f_{xxy}=2\text{and}{f}_{yyx}=0$.

Thus, $f_{xxy}\ne f_{yyx}$.

Therefore, the statement is false.

(d)

To determine

Whether the gradient $\nabla f\left(a,b\right)$ lies in the plane tangent to the surface at $\left(a,b,f\left(a,b\right)\right)$.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Given:

The surface is $z=f\left(x,y\right)$.

Theorem used:

The Gradient and Level Curves:

“Given a function f differentiable at $\left(a,b\right)$, the line tangent to the level curve of f at $\left(a,b\right)$ is orthogonal to the gradient $\nabla f\left(a,b\right)$, provided $\nabla f\left(a,b\right)\ne 0$”.

Description:

The given surface is $z=f\left(x,y\right)$.

By above theorem, it can be concluded that the line tangent to the level curve of f at $\left(a,b\right)$ is orthogonal to the gradient $\nabla f\left(a,b\right)$.

Thus, it does not satisfy the given statement. Because, it is given that the gradient $\nabla f\left(a,b\right)$ lies in the plane tangent to the surface at $\left(a,b,f\left(a,b\right)\right)$.

Here, $\nabla f\left(a,b\right)$ lies in the xy-plane.

Therefore, the statement is false.