Multivariate Calculus: Learning Objective 3

Solve the system of two linear equations for the variables $u$ and $v$ in terms of $x$ and $y$.

$u = \frac{x-2y}{3}$, $v=\frac{x+y}{3}$

We have a system of equations:

1. $x = u+2v$
2. $y = -u+v$

From equation (ii), we can express $u$ in terms of $v$ and $y$: $u = v-y$.
Substitute this into equation (i): \[ x = (v-y) + 2v = 3v-y \] Solve for $v$: \[ x+y = 3v \implies v = \frac{x+y}{3} \] Now substitute the expression for $v$ back into $u=v-y$: \[ u = \frac{x+y}{3} - y = \frac{x+y-3y}{3} = \frac{x-2y}{3} \]

Use exponentiation to eliminate the logarithms. You can also use logarithm properties, such as $\ln(ab) = \ln(a) + \ln(b)$.

$u = e^x$, $v= e^{y-x}$

We start with the given equations:

1. $x = \ln(u)$
2. $y = \ln(uv)$

From equation (i), we can solve for $u$ by exponentiating both sides: \[ u = e^x \] Now, use the property of logarithms on equation (ii): \[ y = \ln(u) + \ln(v) \] We know from (i) that $\ln(u)=x$, so we can substitute that in: \[ y = x + \ln(v) \] Now solve for $\ln(v)$: \[ \ln(v) = y-x \] Finally, exponentiate both sides to solve for $v$: \[ v = e^{y-x} \]

The Jacobian determinant is given by the determinant of the matrix of partial derivatives: \[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \]

$\frac{3}{2}$

1. Find the partial derivatives: \[ \frac{\partial x}{\partial u} = \frac{3u^2}{2} \quad , \quad \frac{\partial x}{\partial v} = 0 \] \[ \frac{\partial y}{\partial u} = \frac{\partial}{\partial u}(vu^{-2}) = -2vu^{-3} = -\frac{2v}{u^3} \quad , \quad \frac{\partial y}{\partial v} = \frac{1}{u^2} \] 2. Calculate the determinant: \[ J = \begin{vmatrix} \frac{3u^2}{2} & 0 \\ -\frac{2v}{u^3} & \frac{1}{u^2} \end{vmatrix} = \left(\frac{3u^2}{2}\right)\left(\frac{1}{u^2}\right) - (0)\left(-\frac{2v}{u^3}\right) = \frac{3}{2} - 0 = \frac{3}{2} \]

Calculate the four partial derivatives and find the determinant of the resulting matrix.

$-1$

1. Find the partial derivatives: \[ \frac{\partial x}{\partial u} = e^v \quad , \quad \frac{\partial x}{\partial v} = ue^v \] \[ \frac{\partial y}{\partial u} = 0 \quad , \quad \frac{\partial y}{\partial v} = -e^{-v} \] 2. Calculate the determinant: \[ J = \begin{vmatrix} e^v & ue^v \\ 0 & -e^{-v} \end{vmatrix} = (e^v)(-e^{-v}) - (ue^v)(0) = -e^{v-v} - 0 = -e^0 = -1 \]

You need to calculate the determinant of a 3x3 matrix of partial derivatives. Expanding along the third column might be easiest.

2

The Jacobian matrix is: \[ \frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{vmatrix} = \begin{vmatrix} 1 & -1 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{vmatrix} \] To find the determinant, we can expand along the third column: \[ J = 0 \cdot \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \] \[ = 1 \cdot ((1)(1) - (-1)(1)) = 1 \cdot (1+1) = 2 \]

Calculate the determinant of the 2x2 matrix. Consider Clairaut's Theorem regarding mixed partial derivatives.

$f_{xx}f_{yy} - (f_{xy})^2$. This is the discriminant, $D$, used in the Second Derivative Test for local extrema.

The determinant of a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $ad-bc$. \[ \det \begin{pmatrix} f_{xx}& f_{xy} \\ f_{yx}& f_{yy} \end{pmatrix} = f_{xx}f_{yy} - f_{xy}f_{yx} \] By Clairaut's Theorem, if the second partial derivatives are continuous, then the mixed partials are equal, i.e., $f_{xy} = f_{yx}$.
So the determinant becomes: \[ f_{xx}f_{yy} - (f_{xy})^2 \] This expression is the discriminant, often denoted $D(x,y)$, which is the key component of the Second Derivative Test used to classify critical points of a two-variable function.

Find the four second-order partial derivatives ($g_{xx}, g_{yy}, g_{xy}, g_{yx}$) and arrange them in the Hessian matrix. Then compute its determinant.

$\textbf{H}g(x,y) = \begin{bmatrix} 6x & -6y \\ -6y & -4-6x \end{bmatrix}$
$\det(\textbf{H}g(x,y)) = -36x^2 - 24x - 36y^2$

1. Find first partial derivatives: \[ g_x = 3x^2 - 3y^2 \] \[ g_y = -4y - 6xy \] 2. Find second partial derivatives: \[ g_{xx} = \frac{\partial}{\partial x}(3x^2 - 3y^2) = 6x \] \[ g_{yy} = \frac{\partial}{\partial y}(-4y - 6xy) = -4 - 6x \] \[ g_{xy} = \frac{\partial}{\partial y}(3x^2 - 3y^2) = -6y \] 3. Assemble the Hessian matrix: \[ \textbf{H}g(x,y) = \begin{bmatrix} g_{xx} & g_{xy} \\ g_{yx} & g_{yy} \end{bmatrix} = \begin{bmatrix} 6x & -6y \\ -6y & -4-6x \end{bmatrix} \] 4. Compute the determinant: \[ \det(\textbf{H}) = (6x)(-4-6x) - (-6y)(-6y) = -24x - 36x^2 - 36y^2 \]

Find the general Hessian matrix first, then substitute the point $(1,2)$ into the matrix, and finally compute the determinant of the resulting matrix of constants.

$\textbf{H}f(1,2) = \begin{bmatrix} 6 & -2\\ -2 & -480 \end{bmatrix}$
$\det(\textbf{H}f(1,2)) = -2884$

1. Find first partial derivatives: \[ f_x = 3x^2 - 2y \] \[ f_y = -2x - 6y^5 \] 2. Find second partial derivatives: \[ f_{xx} = 6x \] \[ f_{yy} = -30y^4 \] \[ f_{xy} = -2 \] 3. Evaluate second partials at (1,2): \[ f_{xx}(1,2) = 6(1) = 6 \] \[ f_{yy}(1,2) = -30(2)^4 = -30(16) = -480 \] \[ f_{xy}(1,2) = -2 \] 4. Assemble the Hessian matrix at (1,2): \[ \textbf{H}f(1,2) = \begin{bmatrix} 6 & -2 \\ -2 & -480 \end{bmatrix} \] 5. Compute the determinant: \[ \det(\textbf{H}f(1,2)) = (6)(-480) - (-2)(-2) = -2880 - 4 = -2884 \]