Chapter 2, Problem 2.1P

a.

To determine

The partial derivatives of the function $f\left(x,y\right)$ .

Answer to Problem 2.1P

The partial derivatives of the function $f\left(x,y\right)$ are

  $\begin{array}{l}\frac{\partial f}{\partial x}=8x,\\ \frac{\partial f}{\partial y}=6y,\\ \frac{{\partial }^{2}f}{\partial x^2}=8,\\ \frac{{\partial }^{2}f}{\partial y^2}=6,\\ \frac{{\partial }^{2}f}{\partial x\partial y}=0\end{array}$

Explanation of Solution

Given information:

The function,

  $f\left(x,y\right)=4x^2+3y^2$

Since,

  $f\left(x,y\right)=4x^2+3y^2$

Partially differentiate with respect to x and y ,

  $\begin{array}{c}\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\left(4x^2+3y^2\right)\\ =4\times 2x+0\\ =8x\end{array}$

and

  $\begin{array}{c}\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left(4x^2+3y^2\right)\\ =0+3\times 2y\\ =6y\end{array}$

Now

  $\begin{array}{c}\frac{{\partial }^{2}f}{\partial x^2}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\right)\\ =\frac{\partial }{\partial x}\left(8x\right)\\ =8\times 1\\ =8\end{array}$

and

  $\begin{array}{c}\frac{{\partial }^{2}f}{\partial y^2}=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial y}\right)\\ =\frac{\partial }{\partial y}\left(6y\right)\\ =6\times 1\\ =6\end{array}$

Also,

  $\begin{array}{c}\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)\\ =\frac{\partial }{\partial x}\left(6y\right)\\ =0\end{array}$

b.

To determine

The derivative $\frac{dy}{dx}$ using the implicit function theorem.

Answer to Problem 2.1P

  $\frac{dy}{dx}=-\frac{8x}{6y}$

Explanation of Solution

Given information:

The function,

  $f\left(x,y\right)=4x^2+3y^2$

and

  $f\left(x,y\right)=16$

Since,

  $f\left(x,y\right)=4x^2+3y^2$

Then,

  $16=4x^2+3y^2$

Differentiate with respect to x ,

  $\begin{array}{l}0=4\times 2x+3\times 2y\frac{dy}{dx}\\ 8x+6y\frac{dy}{dx}=0\\ 6y\frac{dy}{dx}=-8x\\ \frac{dy}{dx}=-\frac{8x}{6y}\end{array}$

c.

To determine

The value of $\frac{dy}{dx}$ at $x=1,y=2$ .

Answer to Problem 2.1P

  $\frac{dy}{dx}=-\frac{2}{3}$

Explanation of Solution

Since,

  $\frac{dy}{dx}=-\frac{8x}{6y}$

Substituting $x=1,y=2$ ,

  $\frac{dy}{dx}=-\frac{8\times 1}{6\times 2}$

  $\begin{array}{c}\frac{dy}{dx}=-\frac{8}{12}\\ =-\frac{2}{3}\end{array}$

d.

To determine

To draw:The graph of $\frac{dy}{dx}=-\frac{2}{3}$ on the surface $f\left(x,y\right)=4x^2+3y^2$ .

Explanation of Solution

Since, $\frac{dy}{dx}$ is the slope of the curve,

From the above part,

  $\frac{dy}{dx}=-\frac{8}{3}$

So, draw the tangent plane on the surface

  $f\left(x,y\right)=4x^2+3y^2$