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PAGE 1

Finding the Equation of a Plane

Find an equation of the plane that contains the point \((2, 1, 1)\) and the line

\[ x = 1 + 3t, \quad y = 2 + t, \quad z = 4 + t \]

Multiple Choice Options

  1. A. \(3x + y + z = 8\)
  2. B. \(2x + y + z = 6\)
  3. C. \(x + 2y + 4z = 8\)
  4. D. \(x - 5y + 2z = -1\) (Selected)
  5. E. \(x - 2y + z = 1\)

Plane Equation Definition

\[ \text{plane: } \vec{n} \cdot \langle x - x_0, y - y_0, z - z_0 \rangle = 0 \]
  • \(\vec{n}\) is the normal vector.
  • \(\langle x - x_0, y - y_0, z - z_0 \rangle\) is a vector in the plane.
  • \((x_0, y_0, z_0)\) is a point in the plane.

Visualizing the Plane and Line

To find the equation, we identify points and vectors from the given information. The point \((2, 1, 1)\) is in the plane. From the line equation, we can extract a point and a direction vector.

\[ \langle 1 + 3t, 2 + t, 4 + t \rangle = \langle 1, 2, 4 \rangle + t \langle 3, 1, 1 \rangle \]
  • One point on the line: \((1, 2, 4)\)
  • Direction vector: \(\langle 3, 1, 1 \rangle\)
A sketch of a parallelogram representing a plane, showing points (2,1,1) and (1,2,4) and vectors <1,-1,-3> and <3,1,1>.
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Determining Vectors in the Plane

We need two vectors in the plane to find the normal vector:

  • \(\vec{u} = \langle 2-1, 1-2, 1-4 \rangle = \langle 1, -1, -3 \rangle\)
  • \(\vec{v} = \langle 3, 1, 1 \rangle\) (the direction vector of the line)

The normal vector \(\vec{n}\) is found by the cross product:

\[ \vec{n} = \vec{u} \times \vec{v} \quad \text{or} \quad \vec{v} \times \vec{u} \]
PAGE 2

Calculating the Normal Vector

We calculate the cross product of \(\vec{u}\) and \(\vec{v}\):

\[ \vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -1 & -3 \\ 3 & 1 & 1 \end{vmatrix} = \vec{i}(2) - \vec{j}(10) + \vec{k}(4) = \langle 2, -10, 4 \rangle \]

Deriving the Plane Equation

Using the normal vector \(\langle 2, -10, 4 \rangle\) and the point \((2, 1, 1)\):

\(\text{plane: } \langle 2, -10, 4 \rangle \cdot \langle x - 2, y - 1, z - 1 \rangle = 0\)

\(2(x - 2) - 10(y - 1) + 4(z - 1) = 0\)

\(2x - 4 - 10y + 10 + 4z - 4 = 0\)

\(2x - 10y + 4z = -2\)

Dividing the entire equation by 2 gives the simplified form:

\[ x - 5y + 2z = -1 \]

This matches option D.

PAGE 3

Trajectory and Speed Calculation

An object is traveling on the trajectory given by \( r(t) = \langle 3t^2 - 1, 4t^2 + 5 \rangle, t \geq 0 \). What is the object's speed at the moment when it has traveled a distance of 5 on this trajectory?

Definitions

distance or length: \( s(t) = \int_{a}^{t} |\vec{r}'(u)| du \)

speed: \( |\vec{r}'(t)| \)

Step 1: Calculate Velocity and Speed

\( \vec{r}'(t) = \langle 6t, 8t \rangle \)

\( |\vec{r}'(t)| = \sqrt{36t^2 + 64t^2} = 10t \)

Step 2: Find time \( t \) for distance of 5

find \( t \) that corresponds to "having traveled distance of 5"

\[ s(t) = \int_{0}^{t} 10u \, du = 5u^2 \Big|_0^t = 5t^2 \]

distance \( = 5 \rightarrow 5t^2 = 5 \rightarrow t = 1 \)

Step 3: Calculate Speed at that time

speed at that time: \( 10(1) = 10 \)

Multiple Choice Options

  • A. \( \frac{1}{10} \)
  • B. \( \frac{1}{5} \)
  • C. 5
  • D. 10 (Correct)
  • E. 50
PAGE 4

Multivariable Limits

The limit \[ \lim_{(x,y) \to (0,0)} \frac{x^4 + y^4}{x^2 + y^2} \] is equal to

  • A. 1
  • B. 0 (Correct)
  • C. 1/2
  • D. 2
  • E. Does not exist
A 2D coordinate system with arrows pointing to the origin from various paths: x=a, y=b, and a diagonal line.

Strategy

limit: show that limit is path dependent (in which case limit DNE)

OR

find limit by path independent way

Checking Simple Paths

along \( x=0 \):
\( \lim_{y \to 0} \frac{y^4}{y^2} = \lim_{y \to 0} y^2 = 0 \)
along \( y=0 \):
\( \lim_{x \to 0} \frac{x^4}{x^2} = \lim_{x \to 0} x^2 = 0 \)
along \( y=x \):
\( \lim_{x \to 0} \frac{x^4 + x^4}{x^2 + x^2} = \lim_{x \to 0} \frac{2x^4}{2x^2} = \lim_{x \to 0} x^2 = 0 \)

Still not enough to prove the limit is 0

time to think of a path independent way

PAGE 5

Multivariable Limits: Algebraic Manipulation

Back to the limit:

\[ \lim_{(x,y) \to (0,0)} \frac{x^4 + y^4}{x^2 + y^2} \]

If the numerator were \( x^4 - y^4 \), then it would be \( (x^2 - y^2)(x^2 + y^2) \), but it's not.

However, the numerator looks a lot like \( (x^2 + y^2)^2 \). Note that:

\[ (x^2 + y^2)^2 = x^4 + 2x^2y^2 + y^4 \]

So, we can rewrite the numerator as:

\[ x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 \]

Evaluating the Limit

\[ \lim_{(x,y) \to (0,0)} \frac{(x^2 + y^2)^2 - 2x^2y^2}{x^2 + y^2} \]
\[ = \lim_{(x,y) \to (0,0)} \left( \frac{(x^2 + y^2)^2}{x^2 + y^2} - \frac{2x^2y^2}{x^2 + y^2} \right) \]

The first part, \( \frac{(x^2 + y^2)^2}{x^2 + y^2} = x^2 + y^2 \), goes to 0.

For the second part, consider the behavior as \( (x,y) \to (0,0) \):

  • Numerator: \( 2x^2y^2 \approx (\text{small \#})^4 \)
  • Denominator: \( x^2 + y^2 \approx (\text{small \#})^2 \)

Since \( (\text{small \#})^4 \) goes to 0 faster than \( (\text{small \#})^2 \), the second part also goes to 0.

\[ = 0 \]
PAGE 6

Partial Derivatives: Mixed Second-Order

If \( f(x,y) = \ln(xy^2 + x) \), find \( f_{xy} \).

Find \( f_{xy} \) for \( f(x,y) = \ln(xy^2 + x) \)

Step 1: Find \( f_x \) (Treating \( y \) as constant)

\[ f_x = \frac{1}{xy^2 + x} \cdot \frac{\partial}{\partial x}(xy^2 + x) \]
\[ = \frac{1}{xy^2 + x} \cdot (y^2 + 1) \]
\[ f_x = \frac{y^2 + 1}{xy^2 + x} \]

Step 2: Find \( f_{xy} \) (Treating \( x \) as constant)

Using the Quotient Rule:

\[ f_{xy} = \frac{(xy^2 + x) \frac{\partial}{\partial y}(y^2 + 1) - (y^2 + 1) \frac{\partial}{\partial y}(xy^2 + x)}{(xy^2 + x)^2} \]
\[ = \frac{(xy^2 + x)(2y) - (y^2 + 1)(2xy)}{(xy^2 + x)^2} \]
\[ = \frac{2xy^3 + 2xy - 2xy^3 - 2xy}{(xy^2 + x)^2} \]
\[ = 0 \]

Correct Option: A (0)

PAGE 7

Directional Derivative Minimum Value

If \( f(x, y) = x^2 + 3y^2 \), for which unit vector \( \vec{u} \) does the directional derivative \( (D_{\vec{u}}f)(2, 1) \) have a minimum value?

directional deriv: \( \vec{\nabla}f \cdot \vec{u} \)

  • \( \vec{\nabla}f \) is the gradient
  • \( \vec{u} \) is the unit vector giving direction

gradient is in direction of greatest increase in \( f(x, y) \)
(greatest ascent or max directional deriv)

min directional deriv: opposite to direction of gradient

\( 0 \) directional deriv \( \rightarrow \) go \( \perp \vec{\nabla}f \) (along a level curve)

  1. A. \( 4\vec{i} + 6\vec{j} \)
  2. B. \( -4\vec{i} - 6\vec{j} \)
  3. C. \( \frac{2}{\sqrt{13}}\vec{i} + \frac{3}{\sqrt{13}}\vec{j} \)
  4. D. \( \frac{-2}{\sqrt{13}}\vec{i} - \frac{3}{\sqrt{13}}\vec{j} \)
  5. E. \( \frac{2}{\sqrt{5}}\vec{i} + \frac{1}{\sqrt{5}}\vec{j} \)

Calculation

\[ \vec{\nabla}f = \langle f_x, f_y \rangle = \langle 2x, 6y \rangle \]\[ \vec{\nabla}f(2, 1) = \langle 4, 6 \rangle \]

make unit vector:

\[ \frac{\langle 4, 6 \rangle}{|\langle 4, 6 \rangle|} = \frac{\langle 4, 6 \rangle}{\sqrt{52}} \]\[ \langle \frac{4}{\sqrt{52}}, \frac{6}{\sqrt{52}} \rangle = \langle \frac{4}{2\sqrt{13}}, \frac{6}{2\sqrt{13}} \rangle = \langle \frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \rangle \]

So opposite is \( \langle \frac{-2}{\sqrt{13}}, \frac{-3}{\sqrt{13}} \rangle \)

PAGE 8

Multivariable Chain Rule

Let \( f(x, y) = 2x^2y + xy^3 \) and \( x = g(s, t) \), \( y = h(s, t) \) are functions of \( s \) and \( t \). Suppose \( g(1, 2) = 1 \), \( h(1, 2) = -1 \) and \( \frac{\partial g}{\partial t}(1, 2) = 2 \), \( \frac{\partial h}{\partial t}(1, 2) = 1 \). Then at \( (s, t) = (1, 2) \), \( \frac{\partial f}{\partial t} \) equals:

Tree diagram for chain rule: f branches to x and y; x branches to s and t; y branches to s and t.
  1. A. 0
  2. B. -10
  3. C. 10
  4. D. 5
  5. E. -5

Chain Rule Formula

\[ \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} \]

evaluate at \( s=1, t=2 \):

\( x(1, 2) = g(1, 2) = 1 \) (given)

\( y(1, 2) = h(1, 2) = -1 \) (given)

\[ \frac{\partial f}{\partial x} = 4xy + y^3 = -4 - 1 = -5 \]\[ \frac{\partial f}{\partial y} = 2x^2 + 3xy^2 = 2 + 3 = 5 \]

Final Calculation:

\[ \frac{\partial f}{\partial t} = (-5)(2) + (5)(1) = -5 \]
PAGE 9

Calculus Review Topics

Other Topics to Review:

  • Surface identification (need to know their names)
  • Level curves

Differentiation and Geometry

Implicit differentiation: chain rule

Curvature

\[ \kappa = \frac{|\vec{T}'(t)|}{|\vec{r}'|} = \frac{|\vec{r}'' \times \vec{r}'|}{|\vec{r}'|^3} \]