Vector Functions

Some of my calc 3 notes. Some equations might not show on mobile.

13.1 Vector-Valued Functions

Vector-Valued Functions

$r(t)$ is a vector that points from the origin to a particles path at a time $t$

$$r(t)=\langle x(t), y(t), z(t) \rangle = x(t)i+y(t)j+z(t)k$$

• In the equation above, variable t is a parameter as t is often representative of time
• $x(t)$, $y(t)$ and $z(t)$ are the components or coordinate functions
• $r(t)$ is referred to as the vector parametrization of a path

Example :: Path vs. Curve:

Describe the path $r(t)= \langle cos(t), sin(t), 1 \rangle, \quad - \infty < t<\infty$

1. Notice z is a constant and cos and sin in x and y respectively. This can be noted that it will form a circle at the height of $z=1$
   

Plane Curve

A line in $\mathbb{R}^2$

Space Curve

Definition: A line (curve) in $\mathbb{R}^3$

Example :: Projecting from $(\mathbb{R}^3)$ onto $(\mathbb{R}^2)$:

Describe the curve traced by $r(t)= \langle cos(t), sin(t), t \rangle, \quad t \le 0$ for in terms of its projections onto the coordinate planes

1. Onto xy-plane: set $z=0$; $r(t)= \langle cos(t), sin(t), 0 \rangle, \quad t \le 0$ = Circle
2. Onto xz-plane: set $y=0$; $r(t)= \langle cos(t), 0, t \rangle, \quad t \le 0$ = Swirl
3. Onto yz-plane: set $x=0$; $r(t)= \langle 0, sin(t), t \rangle, \quad t \le 0$ = Swirl

Example :: Find point on parametrization:

1. $r(4)$ will get us the proper x, therefore I will plug 4 into all values of t
2. $r(4)=(1+4, 2+4^2,4^4)=(5, 18, 256)$ therefore point P does lie on the line

Example :: Find circular parametrization:

Select the correct sine and cosine parametrization $r(t)$ of the intersection of the surfaces $x^2 + y^2 =16$ and $z=3x^2$

1. Knowing we are building a circle based on equation $x^2 + y^2 =R^2$ we can assume the equation w/ the radius $x=4cos(t), y=4sin(t)$ meaning we can write the equation as $r(t)= \langle 4cos(t), 4sin(t), z \rangle$
2. We also know $z=3x^2$, therefore we replace z with this equation getting
3. $r(t)= \langle 4cos(t), 4sin(t), 3x^2 \rangle$
4. $r(t)= \langle 4cos(t), 4sin(t), 48cos^2(t) \rangle$

Example :: Finding plane using parametrizations:

Let $C$ the curve be parameterized by $𝐫(𝑡)=⟨𝑡^2−1,𝑡−2𝑡^2,4−6𝑡⟩$. Evaluate $𝐫(𝑡)$ at $𝑡=0$, $𝑡=1$, and $𝑡=3$.

1. $v_1=r(0)=⟨−1,0,4⟩$
2. $v_2=r(1)=⟨0,-1,-2⟩$
3. $v_3=r(3)=⟨8,-15,-14⟩$

Find an equation for the plane containing $r$ at $𝑡=0$, $𝑡=1$, and $𝑡=3$.

1. Find a point $P_0$ where that I can multiply by the orthogonal vector using the equation $n \cdot P_0$
2. Find the vectors connecting the points of $v_1, v_2, v_3$ as shown above
   1. $v_1=(0,-1,-2)-(-1,0,4)=\langle 1,-1,-6 \rangle$
   2. $v_2=(8, -15, -14)-(-1,0,4)=\langle 9, -15, -18 \rangle$
3. Find the orthogonal vector through the cross product
   1. $$n=v \times w =\begin{bmatrix} i & j & k \\ 1 & -1 & -6 \\ 9 & -15 & -18\end{bmatrix} = i\begin{bmatrix} -1 & -6\ -15 & -18 \end{bmatrix} -j\begin{bmatrix} 1 & -6 \\ 9 & -18 \end{bmatrix} +k \begin{bmatrix} 1 & -1 \\ 9 & -15 \end{bmatrix}$$
      
4. $=\langle -72,-36,-6 \rangle=\langle -12, 6,1 \rangle$
5. Now find point $P_0$ by subtracting 2 vectors
   1. $⟨8,-15,-14⟩-⟨0,-1,-2⟩=(8, -14, -12)$
6. Plug in
   1. $12(-1)+6(0)+1(4)=-8$
   2. (12(𝑡^2−1)+6(𝑡−2𝑡^2)+4−6𝑡=−8)

Determine whether $r_1(𝑡)=⟨𝑡^2+2,𝑡+1,5𝑡−1⟩$ and $r_2(𝑡)=⟨3𝑡,2𝑡−2,𝑡^2−6⟩$ collide or intersect

1. They collide if $r_1(t)=r_2(t)$ where a value of t exists
2. $t+1=2t-2$ equals $t=3$ but when plugged in doesn’t equal out
3. If they don’t collide, they can still intersect. We can find this out if we set $r_1(s)=r_2(t)$ which implies $t=1$ and $s=-1$, which satisfies the equations meaning the equations intersect, but don’t collide at the same moment

The intersection of the plane $𝑦=2$ with the sphere $𝑥^2+𝑦^2+𝑧^2=104$

1. To find this we must plug the values into the equation, and since we know y has a radius of 2 we can create the equation $x^2+2^2+z^2=104$ where $x^2+z^2=100$
2. x=\sqrt{100}cos(t) \quad z=\sqrt{100}sin(t)) therefore \(r(t)= \langle 10cos(t), 2, sin(t) \rangle

13.2 Calculus of Vector-valued Functions

Limit of a Vector-valued Function

A vector-valued function $r(t)$ approaches the limit u (a vector) as t approaches $t_0$ if $\lim_{t\rightarrow t_0} ||r(t)-u|| = 0$ In this case, we write…
$$\lim_{t\rightarrow t_0} r(t)=u$$

Theorem 1: Vector-valued Limits are Computed Component-wise

A vector-valued function $r(t)= \langle x(t), y(t), z(t) \rangle$ approaches a limit as $t \rightarrow t_0$ if and only if each component approaches a limit
$$\lim_{t \rightarrow t_0} r(t)= \langle \lim_{t \rightarrow t_0}x(t), \lim_{t \rightarrow t_0}y(t), \lim_{t \rightarrow t_0}z(t) \rangle$$

Example :: Taking the limit of a vector-value function

Calculate $\lim _{t \rightarrow 3} r(t)), where (r(t)= \langle t^2, 1-t, t^{-1} \rangle$

1. Take the limit of each item individually as shown in… $\lim_{t \rightarrow t_0} r(t)= \langle \lim_{t \rightarrow t_0}x(t), \lim_{t \rightarrow t_0}y(t), \lim_{t \rightarrow t_0}z(t) \rangle$
   
2. $$\lim_{t \rightarrow t_0} r(t)= \langle \lim_{t \rightarrow 3}t^2, \lim_{t \rightarrow 3}(1-t), \lim_{t \rightarrow 3}(\frac{1}{t}) \rangle= \langle 9, -2, 1/3 \rangle$$

Continuous Vector-value Function

A vector-value function is continuous if…
$$\lim_{t \rightarrow t_0}r(t)=t(t_0)$$
Derivative of Theorem 1:
$r(t)$ is continuous at $t_0$ if and only if the components $x(t),y(t),z(t)$ are continuous at $t_0$
$$r'(t)=\frac{d}{dt}r(t)=\lim_{h \rightarrow 0}\frac{r(t+h)-r(t)}{h}$$
$$\lim_{h \rightarrow 0}\frac{r(t+h)-r(t)}{h}=\lim_{h \rightarrow 0} \langle \frac{x(t+h)-x(t)}{h},\frac{y(t+h)-y(t)}{h},\frac{z(t+h)-z(t)}{h} \rangle$$

• If the limit exists as shown above, (r(t)) is differentiable

Theorem 2: Vector-valued Derivatives are computed component-wise

A vector-valued function $r(t)= \langle x(t), y(t), z(t) \rangle$
$$r'(t)= \frac{d}{dt}r(t)=\langle x'(t), y'(t), z'(t) \rangle$$

Example :: Derivative of vector-value function:

Compute the derivative of $r(t)= \langle t,t^4,t^3 \rangle$

1. Take the derivatives of each part separately…
   $$\frac{d}{dt}r(t)=\langle 1,4t^3,3t^2 \rangle$$
   Calculate (r”(3)), where (r(t)= \langle ln(t), t, t^2 \rangle)
2. Solve for $r”(t)) :: (r”(t)= \frac{d}{dt} \langle -t^{-2}, 0, 2 \rangle$
3. Plug in $r”(3)= \langle -1/9, 0, 2 \rangle$

Differentiation Rules

• Sum Rule:
  $$(r_1(t)+r_2(t))’=r_1′(t)+r_2′(t)$$
• Constant Multiple Rule: For any constant
  $$c_1 (cr(t))’=cr'(t)$$
• Scalar Product Rule: For any differentiable scalar-valued function (f_1)
  $$\frac{d}{dt}(f(t)r(t))=f'(t)r(t)+f(t)r'(t)$$
• Chain Rule: For any function of (g),
  $$\frac{d}{dt}r(g(t))=r'(g(t))g'(t)$$

Example :: Using differentiation rules to solve vector-valued problems:

Let $r(t)= \langle t^2, yt, 1 \rangle$ and (f(t)=e^{3t}) Calculate:

1. $\frac{d}{dt}f(t)r(t)$ :: Using scalar product rule $f'(t)r(t)+f(t)r'(t)$
   
2. $$=3e^{3t} \langle t^2, 5t, 1 \rangle + e^{3t} \langle 2t, 5, 0 \rangle=\langle 3e^{3t}(3t^2+2t), 3e^{3t}(15t + 5), 3e^{3t} \rangle$$
   
3. $$\frac{d}{dt}r(f(t)) = r'(f(t))f'(t) = r'(e^{3t}) \cdot 3e^{3t} = \langle 2e^{3t}, 5, 0 \rangle \cdot 3e^{3t} = \langle 6e^{6t}, 15e^{3t}, 0 \rangle$$

Theorem 3: Product rules for Dot and Cross Products

Dot Product Rule:

$$\frac{d}{dt}(r_1(t) \cdot r_2(t))=r’_1(t) \cdot r_2(t)+r_1(t) \cdot r’_2(t)$$

Cross Product Rule:

$$\frac{d}{dt}(r_1(t) \times r_2(t))=r’_1(t) \times r_2(t)+r_1(t) \times r’_2(t)$$

The Derivative of a Tangent Vector

This vector will point in a path tangent to the chosen instant in multiple dimensions, shown as the path traced by (r(t)) at (t=t_0)

$$\frac{\Delta r}{\Delta t}= \frac{r(t_0+h)-r(t_0)}{h}$$

where $h > 0$

Tangent Line

Find the tangent line at $r(t_0)$: $\quad L(t)=r(t_0)+tr'(t_0)$

Example :: parametrization of the tangent line

Find a parametrization of the tangent line at the point indicated
$r(t)=⟨6t,4t^2,2t^3⟩,\quad t=2$

1. Take the derivative of the equation
   
2. $$\int r'(t)=⟨6t,4t^2,2t^3⟩’=⟨6,8t,6t^2⟩$$
3. Solve using the formula: $\quad L(t)=r(t_0)+tr'(t_0)$
   
4. $$L(t)=r(t)+tr'(t)=(6(2),4(2)^2,2(2)^3)+t⟨6,8(2),6(2)^2⟩$$

The Integral of a Tangent Vector

Integrals are taken componentwise as shown below.
$$\int^b_a r(t)dt= \langle \int^b_a x(t)dt,\int^b_a y(t)dt,\int^b_a z(t)dt \rangle$$

Theorem 4:

If $R_1(t)$ and $R_2(t)$ are differentiable and $R’_1(t)=R’_2(t)$, then
$$R_1(t)=R_2(t)+c$$ for some constant $c$

Example :: Taking the integral of a derivative

We must find the integral of $\frac{dr}{dt}=\langle 1-6\sin(3t), \frac{1}{5}t \rangle$ @ $t=4$ and $r(0)=\langle 4,1 \rangle$

1. Find the general solution:
   $$r(t)= \int \langle 1-6\sin(3t), \frac{1}{5}t \rangle=r(t)= \langle t-2\sin(3t), \frac{1}{10}t^2 \rangle +c$$
2. Add these to the equation: $r(0)= \langle 2,0 \rangle +c) and (r(0)= \langle 4,1 \rangle$
   
3. $$r(t)= \langle t+2cos(3t), \frac{1}{10}t^2 \rangle + \langle 2,1 \rangle=r(t)= \langle t+2cos(3t)+2, \frac{1}{10}t^2+1 \rangle$$
4. Then we find $r(4)$

Find $r(t)$ using $r”(𝑡)=⟨e^t,\sin(𝑡),\cos(𝑡)⟩,r(0)=⟨1,0,1⟩,r'(0)=⟨0,4,4⟩$

1. Solve $\int r”(t)$
   $$r'(t)=\int⟨e^t,\sin(𝑡),\cos(𝑡)+1⟩= \langle e^t +c_1, -\cos(t)+c_2, \sin(t)+c_3 \rangle$$
2. Plug in for $r'(0)=⟨0,4,4⟩$
   $$r'(0) = \langle e^t-1, -\cos(t)+5, \sin(t)+4\rangle$$
3. Solve $\int r'(t)$
   $$\int r'(t)=\int⟨e^t-1,-\cos(t)+5,\sin(t)+4⟩= \langle e^t-t+c_1, -\cos(t)+5t+c_2, -\sin(t)+4t+c_3 \rangle$$
4. Plug in for $r(0)=⟨1,0,1⟩$
   $$\langle e^0-0+c_1, -\sin(0)+5(0)+c_2, -\cos(0)+4(0)+c_3 \rangle=…$$
5. $$=\langle e^t-t, -\sin(t)+5t, -\cos(t)+4(t)+2 \rangle$$

Find the location at $𝑡=3$ with $r(0)= \langle 12,13 \rangle$ of a particle whose path satisfies

Integrate: $\frac{dr}{dt}= \langle 18t-\frac{11}{(t+1)^2},2t-4 \rangle$
$$\int \frac{dr}{dt}= r(t) = \langle 9t^2+\frac{11}{(t+1)} +c_1,t^2-4t +c_2 \rangle$$

Plug in for values of (c) and t:
$$r(3)+r(0)= \langle 9(3)^2+\frac{11}{((3)+1)} + 1,(3)^2-4(3) + 13 \rangle= \langle \frac{339}{4},10 \rangle$$

$$= \langle \frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2},\frac{75\pi^2}{16}-\frac{3\pi^2}{16}, \frac{3\pi^2+4}{16}-\frac{\pi^2 + 8}{16}\rangle$$

Fundamental Theorem of Calculus for Vector-valued functions

Part 1: If $r(t)$ is continuous on $[a,b]$, and $R(t)$ is an anti-derivative of $r(t)$, then…
$$\int^b_a r(t)dt=R(b)-R(a)$$
Part 2: Assume that $r(t)$ is continuous on an open interval (I) and let (a) be in (I), then…
$$\frac{d}{dt}\int^t_a r(s)ds=r(t)$$

13.3 Arc Length and Speed

Arc Length

Assuming that $r'(t)$ exists and is a continuous function, then we can calculate length of r over an interval
$$s = \int^b_a ||r'(t) || dt = \int^b_a \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}$$

Example :: Length of a vector

Find the arc length $s$ of the helix given by $r(t)= \langle \cos3t, \sin 3t, 3t \rangle$ for $0 \le t \le 2 \pi$

Calculate the derivative of each component in the equation
$$r'(t) = \langle -3\sin3t, 3\cos3t, 3 \rangle$$

Square each component of (r'(t)) and add them
$$r'(t)^2 = 9\sin^2 3t + 9\cos 3t + 9 = 9(\sin^23t + \cos^2 3t ) + 9 = 18$$

Take the integral
$$s = \int^{2\pi}_0 ||r'(t)||dt = \int^{2\pi}_0 \sqrt{18}dt = 6 \sqrt{2} \pi$$
Compute the length of the curve (𝐫(𝑡)=⟨−8𝑡,5𝑡+6,−5𝑡−2⟩) over the interval (0≤𝑡≤5).

Calculate the derivative of each component
$$r'(t) = ⟨−8,5,−5⟩$$

Square each component, add them and then take the square root
$$||r'(t)|| = \sqrt{(-8)^2 + 5^2 + (-5)^2} = \sqrt{114}$$

Integrate over the range

1. $$s = \int^5_0 \sqrt{114}dt = \sqrt{114}t|^5_0= 5\sqrt{114}$$

Compute the length of the curve (𝐫(𝑡)=⟨2𝑡,ln(𝑡),𝑡^2⟩) over the interval (1≤𝑡≤3).

Calculate the derivative of each component
$$r'(t) = ⟨2,\frac{1}{t},2t⟩$$

Square each component, add them and then take the square root
$$||r'(t)|| = \sqrt{(2)^2 + (\frac{1}{t})^2 + (2t)^2} = \sqrt{4 + \frac{1}{t^2} + 4t^2}$$

$$s = \int^3_0 \sqrt{\frac{4t^2 + 1 + 4t^4}{t^2}}dt = (t^2 + ln|t|)|^3_1 = 9 + ln|3| - 1 - ln|1|$$

Compute the arc length function $s(t)= \int^t_a ||r'(u)||du$ for $r(t)= \langle 14t^2, 12t^2, t^3 \rangle$ where $a=0$

Take the derivative of each component
$$r'(t)= \langle 28t, 24t, 3t^2 \rangle$$

Find the length
$$||r'(t)||=\sqrt{784t^2+576t^2+9t^4}=\sqrt{1360t^2+9t^4}$$

Take the integral

$$\int^t_0 t\sqrt{1360+9t^2}=\frac{(1360+9t^2)^{\frac{3}{2}}}{27}|^t_0=\frac{\left(9t^2+1360\right)^{\frac{3}{2}}-5440\sqrt{85}}{27}$$

Compute the length of the curve $r(t)=5t𝐢+9t𝐣+(6t^2−9)𝐤$ over the interval $0 \le t \le 2$

Find the derivative of each compontent of $r(t)$
$$r'(t)=5i+9j+(12t)k$$

Find the length
$$||r'(t)||=\sqrt{5^2 + 9^2 + (12t)^2}=\sqrt{25 + 81 + 144t^2}=\sqrt{106 + 144t^2}$$

Integrate over the interval
$$\int^2_0 \sqrt{106 + 144t^2}=\sqrt{106 + 144t^2}t|^2_0=\sqrt{106 + 144(2)^2}-\sqrt{106 + 0^2}=\sqrt{682}-\sqrt{106}=15.819$$

Speed of a arc length function

$$s(t) = \frac{ds}{dt}=||r'(t)||$$

Example :: Speed of an arc length function

Find the speed over the path $r(t)=⟨e^{t−4},−7,4t^{−1}⟩$ at $t=4$; or $v(4)$

1. Find the derivative of each compontent
   $$r'(t)=⟨e^{t−4},0,-4t^{−2}⟩$$
2. Find the length
   $$||r'(t)||=v(t)=\sqrt{e^{2t−8} + 16t^{-4}}$$
3. Plug in (t=4)
   $$\sqrt{e^{2(4)−8} + 16(4)^{-4}}=\sqrt{1 + 1/16}$$

Arc Length Parametrization

Note: parametrizations of a curve are often not unique, as shown in $r_1(t) = \langle t, t^2 \rangle$ is equal to $r_2(u)= \langle u^3, u^6 \rangle$
$$||r'(s)||=1 \quad \text{for all s}$$
3 properties of acr length parametrization:

1. Every velocity vector $r'(s)$ has length equal to 1
2. The arc length of the curve that is traced over any interval $[a, b]$ is equal to $b-a$, the length of the interval:
   $$\text{distance traveled over [a,b]} = \int^b_a||r'(s)||ds=\int^b_a1dt=b-a$$
3. The parameter $s$ corresponds to the arc length of the curve that is traced from the starting point, thus the name arc length parametrization

Finding the arc length parametrization

Note: Start with any parametrization $r(t)$ such that $r'(t) \ne 0$ for all $t$.

1. Form the arc length integral
   $$s = g(t)= \int^t_0 ||r'(u)||du$$
2. Determine the inverse of $g(t)$
   • Note that because $||r'(t)|| \ne 0, \quad s = g(t)$ is an increasing function, and $g$ has an inverse $t=g^{-1}(s)$
3. Take the new parametrization to find the arc length parametrization
   $$r_1(s) = r(g^{-1}(s))$$

Example :: Finding the arc length parametrization:

Find the parametrization $𝐫(𝑡)=⟨13cos(𝑡),13sin(t), \frac{33t}{2\pi}⟩$ of the helix.

1. First, calculate the arc length function
   $$||r'(t)||= \sqrt {169\sin^2(t) + 169\cos^2(t) + \frac{33^2}{4\pi^2}}=\sqrt{169 + \frac{1089}{4\pi^2}}$$
   $$\int^t_0\sqrt{169 + \frac{1089}{4\pi^2}}=\sqrt{169 + \frac{1089}{4\pi^2}}t$$
2. Find the inverse of the arc length function
   $$s = \sqrt{169 + \frac{1089}{4\pi^2}}t; \quad t = \frac{s}{\sqrt{169 + \frac{1089}{4\pi^2}}}$$
3. Substitute our new $t$ into the original parametrization gives us the arc length parametrization
   $$r(\frac{s}{\sqrt{169 + \frac{1089}{4\pi^2}}})=⟨13cos(\frac{s}{\sqrt{169 + \frac{1089}{4\pi^2}}}),13\sin(\frac{s}{\sqrt{169 + \frac{1089}{4\pi^2}}}), \frac{33(\frac{s}{\sqrt{169 + \frac{1089}{4\pi^2}}})}{2\pi}⟩$$
4. To verify the answer, see if your new equation equals 1, but I don’t care to do that

13.5 Motion in 3-Space

Velocity vector is the derivative:
$$v(t)=r'(t)=\lim_{h \rightarrow 0} \frac{r(t+h)-r(t)}{h}$$
Acceleration vector is the 2nd derivative:
$$a(t)=r”(t)$$

Theorem 1: Tangential and Normal Components of Acceleration

In the decomposition $a=a_T T + a_N N$, we have
$$a_T=a \cdot T = \frac{a \cdot v}{||v||}, \quad a_N=a \cdot N = \sqrt{||a||^2-|a_T|^2}$$
and…
$$a_T T=(\frac{a \cdot v}{v \cdot v})v, \quad a_N N=a-a_T T= a - (\frac{a \cdot v}{v \cdot v})v$$

Example :: The decomposition of a(t):

Find the decomposition of $a(t)$ where $a_TT+a_NN$ into tangential and normal components for $r(𝑡)=⟨5𝑡,3𝑡^2,3𝑡^3⟩$ at $t=2$

1. Solve for a using double derivative to get to $a(t)$
   $$v(t)=⟨5,6t,9t^2⟩; \quad a(t)=⟨0,6,18t⟩$$
2. Solve for T at (t=2)
   $$v(2)=⟨5,12,36⟩,\quad a(2)=⟨0,6,36⟩$$
   $$T=\frac{v}{||v||}=\frac{\langle 5,12,36 \rangle}{\sqrt{5^2+12^2+36^2}}= \frac{\langle 5,12,36 \rangle}{\sqrt{1465}}$$
3. Solve for $a_T$
   $$a_T=a \cdot T=⟨0,6,36⟩ \cdot \langle \frac{5}{\sqrt{1465}}, \frac{12} {\sqrt{1465}}, \frac{36}{\sqrt{1465}}\rangle=\frac{\langle 72,36^2 \rangle}{\sqrt{1465}}=\frac{93,312 }{\sqrt{1465}}$$
4. Solve for $a_TT$
   $$a_TT=a_T \cdot T= \frac{93,312 }{\sqrt{1465}} \cdot \frac{\langle 5,12,36 \rangle}{\sqrt{1465}} =\langle 63.694, 764.33,2, 292.991 \rangle$$
5. Solve for $a_NN$:
   $$a_NN=a-a_TT=\langle 0,6,36 \rangle-\langle 63.694, 764.33,2, 292.991 \rangle = \langle 63.694, 758.33,2, 256.991 \rangle$$
6. Solve for $a_NN$
   $$||\langle 63.694, 758.33,2, 256.991 \rangle||=2,381.833$$

Homework 4

The position of a particle after t seconds is given by $r(t)= \langle t^2, 4t, 4 \ln t \rangle$ for $t \ge 1$

1. Calculate the position, velocity, and acceleration of the particle when $t=1$
   1.1. position:
   $$r(1)= \langle 1^2, 4(1), 4 \ln (1) \rangle= \langle 1, 4, 0 \rangle$$
   1.2. Velocity: (r'(t)= \langle 2t, 4, \frac{4}{t} \rangle)
   $$r'(1)= \langle 2(1), 4, \frac{4}{1} \rangle= \langle 2, 4, 4 \rangle$$
   1.3. Acceleration: (r”(t)= \langle 2, 0, -\frac{4}{t^2} \rangle)
   $$r”(t)= \langle 2, 0, -4 \rangle$$
2. Calculate the total distance traveled by the particle from $t=1$ to $t=e$
   2.1. Area under the graph (distance): $\int^b_a ||r'(t)||$
   $$||r'(t)||=\langle (2t)^2 + 4^2 + (\frac{4}{t})^2 \rangle=\sqrt{4t^2+16+ \frac{16}{t^2}}$$
   $$\sqrt{4t^2+16+ \frac{16}{t^2}}=\sqrt{\frac{4t^4+16t^2+16}{t^2}}=\sqrt{\frac{4(t^2+2)^2}{t^2}}=\frac{2(t^2+2)}{t}$$
   2.2. Integrate:
   $$2\int ^e _1 \frac{t^2+2}{t}=2(\int ^e _1t+\int ^e _1\frac{2}{t})= t^2+4\ln t |^e_1$$
   2.3. Solve:
3. $$t^2+4\ln t |^e_1=e^2+4\ln e - 1^2+4\ln 1=e^2+4-1=e^2 +3$$

A scientist with a jetpack starts at (t = 0) at the position $(2, 5, 1)$ and flies with a velocity function of $v(t) = \langle 2t −3, 4t, 6e^{−2t} \rangle$, in m/s. He flies for 2 seconds before…

1. Find a formula for the scientist’s position as a function of time
   1.1. $\int v(t) = r(t)$
   $$\int v(t) = \langle t^2 −3t + 2, 2t^2 + 5, -3e−t^2 +4 \rangle$$
2. At what position does the scientist end up after the 2 seconds?
   2.1. Plug in for $t=2$
   $$r(2) = \langle 2^2 −3(2) + 2, 2(2)^2 + 5, -3e−(2)^2 +4 \rangle=v(t) = \langle 0, 13, -3e^{-4} + 4 \rangle$$
3. What is the scientist’s velocity vector when (t = 2)?
   3.1. Solve for the equation given:
4. $$v(2) = \langle 2(2) −3, 4(2), 6e^{−2(2)} \rangle = \langle 1, 8, 6e^{−4} \rangle$$

Oh no! After the 2 seconds, the jetpack ran out of fuel and the scientist begins to fall. The scientist is now only affected by the acceleration due to gravity (which we approximate as (a = \langle 0, 0, −10 \rangle) ).

1. Get a formula for the path the scientist takes as they fall. Use answers to parts (b) and (c) of the previous problem as conditions for the velocity and acceleration when $t = 2$.
2. FINISH!! (Never finished)
3. —
   A roller coaster is going along a path. Wheeee! At a certain time $t_0$ sec, the roller coaster has position (in m) $r(t_0) = 〈5, 2, 13〉$, velocity (in m/s) $v(t_0) = 〈−2, 3, 2〉$ and acceleration (in (m/s^2)) $a(t_0) = 〈7, −3, 5〉$.
4. What is the speed of the roller coaster at this time $(t_0)$
   1.1. Find the length of the velocity vector to find speed
   $$||v(t_0)||=\sqrt{(-2)^2+3^2+2^2}=\sqrt{17} m/s$$
5. Was the roller coaster speeding up or slowing down?
   2.1. Why is the roller coaster slowing down
6. Approximate the position of the roller coaster a quarter of a second later
   3.1. Multiply the velocity by 1/4
7. $$r(t_0)-\frac{1}{4}v(t_0) = (5, 2, 13) + (-0.5, .75, 0.5)=(4.5, 2.75, 13.5)$$

[P] A second roller coaster follows the curve R defined by the following system of equations:

1. Give a parametrization for the path the roller coaster takes
   1.1. Solve for x and y. Since (x^2 + y^2), this means our shape is a circle giving us cos and sin for the first two components
   $$r(t)=\langle 10\cos t, 10 \sin t, 10 \cos^2 t \rangle \quad 0 \le t \le 2\pi$$
2. During which parts of the track is the roller coaster going uphill?
   2.1 Thinking in terms of cos being graphed, we can assume that from (0 \le t \le \frac{\pi}{2}) and (\pi \le t\frac{3\pi}{2})
3. [C] Set up an integral to compute the length of the roller coaster track. Then use a computer to evaluate the integral; give your answer to 3 decimal places.
   3.1. (\int^{2\pi}_0||r(t)||) HELP
   $$||r'(t)||=\sqrt{100\sin^2 (t) + 100 \cos^2 (t) + 100 \sin^4 (2t)}=73.0506$$

Consider the (continuous) function which has some values given by the following table (the horizontal axis is the x-axis and the vertical axis is the y-axis):

1. What is the value of $f(2.5, 4.1)$?
   • ANS: $96$
2. Approximate the value of $f(2.6, 4.1)$.
   • $\left(\frac{96 \cdot 93}{2}\right) \approx 94.5$
3. What is the average rate of change of $f$ moving from the point $(2.5, 4.1)$ to $(2.5, 4.3)$?
   • $\frac{\Delta y}{\Delta x} = \frac{4}{0.2} = 20$
4. What is the average rate of change of $f$ moving from the point $(2.3, 3.9)$ to $(2.7, 4.1)$?
   • $\frac{\Delta y}{\Delta x} = \frac{93-86}{\sqrt{0.4^2+0.2^2}} = \frac{7}{0.4472} \approx 15.65$

1. Consider the function $f(x, y) = \sqrt{x^2 + y^2}$. What does the graph of this function look like?
   • A cone when translated to $z = \sqrt{x^2 + y^2}$
2. Give a parametrization for the line $y = 2x + 3$ and plug the parametrization into $f$. What is the relationship between $f(x, y)$ and $f(t)$?
   • Plug in $y$ for $x$ and change formula to $f(t) = \ldots$
3. $f(x, 2x+3) = f(t) = \sqrt{t^2 + (2t+3)^2} = \sqrt{t^2 + 4t^2 + 12t + 9} = \sqrt{5t^2 + 12t + 9}$
4. Minimize this one-variable function to find which point on the line $y = 2x + 3$ is closest to the origin.