Multivariate Calculus: Learning Objectives 1, 2

The curve is formed by slicing the surface with a plane where $y=y_0$ (constant). A direction vector for the tangent line must reflect this. The slope of this curve is given by the partial derivative with respect to $x$.

(b) $\langle \Delta x, 0, f_x(x_0,y_0) \cdot \Delta x \rangle$

Slicing the surface $z=f(x,y)$ with a vertical plane $y=y_0$ creates a curve in that plane. The slope of the tangent line to this curve is given by the partial derivative with respect to $x$, $f_x(x_0, y_0)$. A direction vector for this tangent line will have components showing change in $x$, $y$, and $z$.

• The change in $y$ is 0, since we are in the plane $y=y_0$.
• Let the change in $x$ be some non-zero value $\Delta x$.
• The corresponding change in $z$ along the tangent line is given by (slope) $\times$ (run), which is $\Delta z = f_x(x_0, y_0) \cdot \Delta x$.

Therefore, a direction vector is $\langle \Delta x, 0, f_x(x_0,y_0) \cdot \Delta x \rangle$.

The gradient vector is defined as $\nabla f = \langle f_x, f_y \rangle = \frac{\partial f}{\partial x}\textbf{i} + \frac{\partial f}{\partial y}\textbf{j}$. Find the partial derivatives with respect to $x$ and $y$.

$\nabla f(x,y) = (2x-y)\textbf{i} + (-x+6y)\textbf{j}$

1. Find the partial derivative with respect to $x$, treating $y$ as a constant: \[ f_x = \frac{\partial}{\partial x}(x^2-xy +3y^2) = 2x-y \] 2. Find the partial derivative with respect to $y$, treating $x$ as a constant: \[ f_y = \frac{\partial}{\partial y}(x^2-xy +3y^2) = -x+6y \] 3. Combine them into the gradient vector: \[ \nabla f(x,y) = \langle f_x, f_y \rangle = \langle 2x-y, -x+6y \rangle \] Or using standard basis vectors: $(2x-y)\textbf{i} + (-x+6y)\textbf{j}$.

The gradient is $\nabla f = \langle f_x, f_y, f_z \rangle$. Calculate the three partial derivatives.

$(10x-2y+3z)\textbf{i} + (-2x+2y-4z)\textbf{j} + (3x-4y+2z)\textbf{k}$

We find each partial derivative: \[ f_x = \frac{\partial}{\partial x}(5x^2 -2xy+y^2-4yz+ z^2+3xz) = 10x - 2y + 3z \] \[ f_y = \frac{\partial}{\partial y}(5x^2 -2xy+y^2-4yz+ z^2+3xz) = -2x + 2y - 4z \] \[ f_z = \frac{\partial}{\partial z}(5x^2 -2xy+y^2-4yz+ z^2+3xz) = -4y + 2z + 3x \] The gradient is: \[ \nabla f = \langle 10x-2y+3z, -2x+2y-4z, 3x-4y+2z \rangle \]

First find the general gradient vector $\nabla f = \langle f_x, f_y, f_z \rangle$. Then substitute the coordinates of the point $(-2,-1,-1)$ into your result.

$\sqrt{2}\textbf{i}+\sqrt{2}\textbf{j}+ \sqrt{2}\textbf{k}$

1. Find the partial derivatives: \[ f_x = \frac{\partial}{\partial x}(x(y^2+z^2)^{1/2}) = (y^2+z^2)^{1/2} = \sqrt{y^2+z^2} \] \[ f_y = x \cdot \frac{\partial}{\partial y}((y^2+z^2)^{1/2}) = x \cdot \frac{1}{2}(y^2+z^2)^{-1/2}(2y) = \frac{xy}{\sqrt{y^2+z^2}} \] \[ f_z = x \cdot \frac{\partial}{\partial z}((y^2+z^2)^{1/2}) = x \cdot \frac{1}{2}(y^2+z^2)^{-1/2}(2z) = \frac{xz}{\sqrt{y^2+z^2}} \] 2. Evaluate the partial derivatives at $(-2,-1,-1)$: \[ f_x(-2,-1,-1) = \sqrt{(-1)^2+(-1)^2} = \sqrt{2} \] \[ f_y(-2,-1,-1) = \frac{(-2)(-1)}{\sqrt{(-1)^2+(-1)^2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \] \[ f_z(-2,-1,-1) = \frac{(-2)(-1)}{\sqrt{(-1)^2+(-1)^2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \] 3. Assemble the gradient vector: \[ \nabla f(-2,-1,-1) = \langle \sqrt{2}, \sqrt{2}, \sqrt{2} \rangle \]

The mixed partial derivative $f_{xy}$ represents the rate of change of the slope in the x-direction ($f_x$) as you move in the y-direction. What does a positive value for this derivative imply?

(c) the slope in the $x$-direction increases as one moves in the positive $y$-direction

The partial derivative $f_x = \frac{\partial f}{\partial x}$ represents the slope of the tangent line in the $x$-direction.

The mixed partial derivative $f_{xy} = \frac{\partial}{\partial y}(f_x)$ represents the rate of change of $f_x$ with respect to $y$.

Since we are given that $f_{xy}(x_0,y_0) > 0$, this means that the slope in the x-direction, $f_x$, is increasing as one moves in the positive y-direction from the point $(x_0, y_0)$.

Critical points occur where both partial derivatives, $g_x$ and $g_y$, are equal to zero. Find these derivatives and solve the resulting system of linear equations.

The only critical point is $(-1,-1)$.

1. Find the partial derivatives: \[ g_x = \frac{\partial}{\partial x}(x^2+2xy-4y^2+4x-6y+4) = 2x+2y+4 \] \[ g_y = \frac{\partial}{\partial y}(x^2+2xy-4y^2+4x-6y+4) = 2x-8y-6 \] 2. Set the derivatives to zero and solve the system:

1. $2x+2y+4=0 \implies x+y = -2$
2. $2x-8y-6=0 \implies x-4y = 3$

Subtract equation (ii) from (i): \[ (x+y) - (x-4y) = -2 - 3 \] \[ 5y = -5 \implies y = -1 \] Substitute $y=-1$ into equation (i): \[ x + (-1) = -2 \implies x = -1 \] The only critical point is $(-1, -1)$.

Use the Second Derivative Test. You need to calculate the discriminant $D(x,y) = f_{xx}f_{yy} - [f_{xy}]^2$ and evaluate it at the critical point $(1,1)$.

The point $(1,1)$ is a relative minimum.

1. Find the second partial derivatives: \[ f_{xx} = \frac{\partial}{\partial x}(4x^3-4y) = 12x^2 \] \[ f_{yy} = \frac{\partial}{\partial y}(4y^3-4x) = 12y^2 \] \[ f_{xy} = \frac{\partial}{\partial y}(4x^3-4y) = -4 \] 2. Calculate the discriminant D: \[ D(x,y) = f_{xx}f_{yy} - [f_{xy}]^2 = (12x^2)(12y^2) - (-4)^2 = 144x^2y^2 - 16 \] 3. Evaluate D at the critical point (1,1): \[ D(1,1) = 144(1)^2(1)^2 - 16 = 144 - 16 = 128 \] 4. Apply the Second Derivative Test: Since $D(1,1) = 128 > 0$, we check the sign of $f_{xx}(1,1)$. \[ f_{xx}(1,1) = 12(1)^2 = 12 > 0 \] Because $D > 0$ and $f_{xx} > 0$, the function has a relative minimum at $(1,1)$.

Find the critical point by setting both first partial derivatives to zero. Then apply the Second Derivative Test using $D = f_{xx}f_{yy} - [f_{xy}]^2$.

$f$ has a local minimum at $(-1,2)$. The minimum value is -16.

1. Find critical points: \[ f_x = 8x+8 = 0 \implies x=-1 \] \[ f_y = 18y-36 = 0 \implies y=2 \] The only critical point is $(-1,2)$.

2. Find second derivatives: \[ f_{xx} = 8 \] \[ f_{yy} = 18 \] \[ f_{xy} = 0 \] 3. Apply the Second Derivative Test: \[ D = (8)(18) - (0)^2 = 144 > 0 \] Since $D > 0$ and $f_{xx} = 8 > 0$, the function has a local minimum at $(-1,2)$.
The minimum value is $f(-1,2) = 4(-1)^2 + 9(2)^2 + 8(-1) - 36(2) + 24 = 4 + 36 - 8 - 72 + 24 = -16$.

First, find any critical points inside the rectangular domain. Then, check the function's values along the four boundary edges of the rectangle. Finally, compare the function values at the critical point and at all points found on the boundary.

The absolute maximum is 36, which occurs at $(0,2)$. The absolute minimum is 17, which occurs at the critical point $(3,1)$.

1. Find critical points: \[ f_x = 2x - 2y - 4 = 0 \implies x-y=2 \] \[ f_y = -2x + 8y - 2 = 0 \implies -x+4y=1 \] Adding the two equations: $3y=3 \implies y=1$. Then $x-1=2 \implies x=3$. The critical point is $(3,1)$, which is inside the domain. $f(3,1) = 9-6+4-12-2+24 = 17$.

2. Check the boundary edges:

• Edge 1 ($y=0, 0 \le x \le 4$): $f(x,0) = x^2-4x+24$. $f'(x,0)=2x-4=0 \implies x=2$. Check endpoints and this point: $f(0,0)=24, f(4,0)=24, f(2,0)=20$.
• Edge 2 ($y=2, 0 \le x \le 4$): $f(x,2) = x^2-4x+16-4x-4+24 = x^2-8x+36$. $f'(x,2)=2x-8=0 \implies x=4$. Check endpoints: $f(0,2)=36, f(4,2)=16-32+36=20$.
• Edge 3 ($x=0, 0 \le y \le 2$): $f(0,y) = 4y^2-2y+24$. $f'(0,y)=8y-2=0 \implies y=1/4$. Check endpoints and this point: $f(0,0)=24, f(0,2)=36, f(0,1/4)=23.75$.
• Edge 4 ($x=4, 0 \le y \le 2$): $f(4,y) = 16-8y+4y^2-16-2y+24 = 4y^2-10y+24$. $f'(4,y)=8y-10=0 \implies y=5/4$. Check endpoints and this point: $f(4,0)=24, f(4,2)=20, f(4,5/4)=17.75$.

3. Compare all values: The function values we found are: 17 (critical point), 20, 24, 36, 23.75, 17.75. The largest is 36 and the smallest is 17.
The absolute maximum is 36 at $(0,2)$ and the absolute minimum is 17 at $(3,1)$.

Use the method of Lagrange multipliers. Minimize the squared distance function $f(x,y,z) = x^2+y^2+z^2$ subject to the constraint $g(x,y,z) = x^2-yz-5=0$. Solve the system $\nabla f = \lambda \nabla g$.

$(\sqrt{5},0,0)$ and $(-\sqrt{5},0,0)$.

We want to minimize $f(x,y,z)=x^2+y^2+z^2$ subject to $g(x,y,z)=x^2-yz-5=0$.
1. Find gradients: $\nabla f = \langle 2x, 2y, 2z \rangle$
$\nabla g = \langle 2x, -z, -y \rangle$
2. Set up Lagrange equations $\nabla f = \lambda \nabla g$:

1. $2x = \lambda(2x) \implies 2x(1-\lambda)=0$
2. $2y = \lambda(-z)$
3. $2z = \lambda(-y)$

3. Solve the system: From (i), we have two cases: $x=0$ or $\lambda=1$.
Case 1: $\lambda=1$. From (ii), $2y = -z$. From (iii), $2z = -y$. Substituting $z=-2y$ into the second equation gives $2(-2y)=-y \implies -4y=-y \implies 3y=0 \implies y=0$. If $y=0$, then $z=0$. We plug these into the constraint: $x^2 - (0)(0) = 5 \implies x^2=5 \implies x=\pm\sqrt{5}$. This gives points $(\sqrt{5},0,0)$ and $(-\sqrt{5},0,0)$. The squared distance is $f(\pm\sqrt{5},0,0) = 5$.
Case 2: $x=0$. The constraint becomes $-yz=5 \implies y=-5/z$. From (ii) and (iii), we get $2y=-\lambda z$ and $2z=-\lambda y$. Multiplying these gives $4yz = \lambda^2 yz$. Since $yz=-5 \neq 0$, we can divide by $yz$ to get $\lambda^2=4 \implies \lambda = \pm 2$. If $\lambda=2$, then $2y=-2z \implies y=-z$. Substituting into $-yz=5$ gives $-(-z)z=5 \implies z^2=5$. So $z=\pm\sqrt{5}$ and $y=\mp\sqrt{5}$. This gives points $(0, -\sqrt{5}, \sqrt{5})$ and $(0, \sqrt{5}, -\sqrt{5})$. The squared distance is $f = 0^2+5+5=10$.
Comparing the squared distances (5 and 10), the minimum distance is $\sqrt{5}$ and it occurs at the points $(\sqrt{5},0,0)$ and $(-\sqrt{5},0,0)$.