MVC PS#18

We will first establish an equation for the statement w.r.t. \(z\):

\begin{align} &x^2 - 3y^2 +z^2 = 7 \\ \Rightarrow &z^2 = 7-x^2 + 3y^2 \\ \Rightarrow &z = \sqrt{7-x^2 + 3y^2} \end{align}

Then, we find the components of the slope in each of the dimensions.

\begin{align} \frac{\partial}{\partial x} \sqrt{7-x^2 + 3y^2} \\ \Rightarrow \frac{-2x}{2\sqrt{7-x^2 + 3y^2}} \\ \Rightarrow \frac{-x}{\sqrt{7-x^2 + 3y^2}} \end{align} \begin{align} \frac{\partial}{\partial y} \sqrt{7-x^2 + 3y^2} \\ \Rightarrow \frac{6y}{2\sqrt{7-x^2 + 3y^2}} \\ \Rightarrow \frac{3y}{\sqrt{7-x^2 + 3y^2}} \end{align}

Therefore, at point \((1,1,3)\), the gradient is as follows

\begin{equation} \begin{bmatrix} \frac{-1}{3} \\ 1 \end{bmatrix} \end{equation}

The basic equation for the plane, then, would be:

\begin{equation} z = \frac{-1}{3}x+1y + b \end{equation}

As the plane is tangent to \((1,1,3)\), we will supply these values and calculate the necessary constant.

\begin{align} &3 = -\frac{1}{3}+1 + b \\ \Rightarrow &3 = \frac{2}{3} + b \\ \Rightarrow &b = \frac{7}{3} \end{align}

We first will figure the first and seconds derivatives of this function to check and verify its critical points.

\begin{equation} f'(x) = 4x^3 - 18x \end{equation} \begin{equation} f''(x) = 12x^2 - 18 \end{equation}

To figure the critical points (maxima, minima, inflection), we first solve for the points in which \(f'(x)\) is \(0\).

\begin{align} &0 = 4x^3 - 18x \\ \Rightarrow & 0 = x(4x^2 - 18) \\ \Rightarrow & x = \{0, \frac{3\sqrt{2}}{2}, \frac{-3\sqrt{2}}{2}\} \end{align}

We then figure the second derivatives of the function at these two points.

\begin{equation} f''(\{0, \frac{3\sqrt{2}}{2}, \frac{-3\sqrt{2}}{2}\}) = \{-18, 36, 36 \} \end{equation}

At the first critical point, the second derivative is negative. By the second-derivative test, therefore, this point serves as a local maxima is the function, after that point, would begin to exhibit a negative slope and then proceed to decrease.

At the second and third critical points, the second derivative is positive. By the second-derivative test, therefore, this point serves as local minima as the function, after that point, would begin to exhibit a positive slope and then proceed to increase.

The coordinates of the critical points are, therefore, as follows:

Type	x	y
maxima	0	0
minima	3sqrt(2)/2	-20.25
minima	-3sqrt(2)/2	-20.25

We first will figure the first and seconds derivatives of this function to check and verify its critical points.

\begin{equation} f'(x) = 5x^4 - 4x^3 - 18x^2 \end{equation} \begin{equation} f''(x) = 20x^3 - 12x^2 - 36x \end{equation}

To figure the critical points (maxima, minima, inflection), we first solve for the points in which \(f'(x)\) is \(0\).

\begin{align} &0 = 5x^4 - 4x^3 - 18x^2 \\ \Rightarrow &0 = x^2 (5x^2 - 4x^1 - 18) \\ \Rightarrow &x = \{0, \frac{4 + \sqrt{376}}{10}, \frac{4 - \sqrt{376}}{10}\} \end{align}

We then figure the second derivatives of the function at these two points.

\begin{equation} f''(\{0, \frac{4 + \sqrt{376}}{10}, \frac{4 - \sqrt{376}}{10}\}) \approx \{0, 106.09, -45.93\} \end{equation}

At the first critical point, the second derivative is zero. By the second-derivative test, this would mean that that point is an inflection point — the function changes increasing/decreasing status at that point..

At the second critical point, the second derivative is positive. By the second-derivative test, therefore, this point serves as local minima as the function, after that point, would begin to exhibit a positive slope and then proceed to increase.

At the third critical point, the second derivative is negative. By the second-derivative test, therefore, this point serves as a local maxima is the function, after that point, would begin to exhibit a negative slope and then proceed to decrease.

The coordinates of the critical points are, therefore, as follows:

Type	x	y
inflection	0	0
minima	(4+sqrt(346))/10	-36.7
maxima	(4-sqrt(346))/10	7.63