Parametric, Polar, and Vector Functions - AP Calculus BC

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Question

Find the vector form of to .

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Answer

When we are trying to find the vector form we need to remember the formula which states to take the difference between the ending and starting point.

Thus we would get:

Given and

$\overrightarrow{v}=[d-a, e-b, f-c]$

In our case we have ending point at and our starting point at .

Therefore we would set up the following and simplify.

$\overrightarrow{v}=[6-0,3-1,1-3]=[6,2,-2]$

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Question

$\newline {\vec{r}(t)} = $(5t^3$$,e^{11t}$$,e^{5t}$)\newline \textnormal{Calculate } $\frac{\mathrm{d}$ }{\mathrm{d} t}{\vec{r}(t)}$

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Answer

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline $\frac{\mathrm{d}$ }{\mathrm{d} t}{\vec{r}(t)} = ($\frac{\mathrm{d}$ }{\mathrm{d} t}f(t), $\frac{\mathrm{d}$ }{\mathrm{d} t}g(t),$\frac{\mathrm{d}$ }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d}$ }{\mathrm{d} t} $t^n$ = n * $t^{n-1}$

Special rule when differentiating an exponential function: $\frac{\mathrm{d}$ }{\mathrm{d} t} $e^{kt}$ = $ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = $(5t^3$$,e^{11t}$$,e^{5t}$)$

$\frac{\mathrm{d}$ }{\mathrm{d} $t}5t^3$$=15t^2$

$\frac{\mathrm{d}$ }{\mathrm{d} $t}e^{11t}$$=11e^{11t}$

$\frac{\mathrm{d}$ }{\mathrm{d} $t}e^{5t}$$=5e^{5t}$

Put it all together to get $$\frac{\mathrm{d}$ }{\mathrm{d} t}{\vec{r}(t)}$

$$\frac{\mathrm{d}$ }{\mathrm{d} t}{\vec{r}(t)} = $(15t^2$$,11e^{11t}$$,5e^{5t}$)$

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Question

$\newline\vec{v} $=(x^2$,-x,5)\newline\vec{u} =(2,3x,11)\newline$

Calculate $\vec{v} + \vec{u}$

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Answer

Calculate the sum of vectors.

In general,

$\vec{v} = (v_1,v_2,v_3)$

$\vec{u} = (u_1,u_2,u_3)$

$\vec{v} + \vec{u} = (v_1 + u_1, v_2 + u_2,v_3 + u_3)$

Solution:

$\newline\vec{v} $=(x^2$,-x,5)\newline\vec{u} =(2,3x,11)\newline$

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Question

Given points and , what is the vector form of the distance between the points?

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Answer

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point

$a=\left \langle $a_{1}$$,a_{2}$$,a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ ,

the distance is the vector

$v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points and , we get:

$v=\left \langle 9-2,1-4,0-5\right \rangle$

$v=\left \langle 7,-3,-5\right \rangle$

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Question

Given points and , what is the vector form of the distance between the points?

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Answer

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point $a=\left \langle $a_{1}$$,a_{2}$$,a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , the distance is the vector $v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points and , we get:

$v=\left \langle 7-4,1-2,0-(-2)\right \rangle$

$v=\left \langle 3,-1,2\right \rangle$

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Question

The graph of the vector function $r(t)=\left \langle $\frac{1}{2}$t,\sin(t)\right \rangle$ can also be represented by the graph of which of the following functions in rectangular form?

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Answer

We can find the graph of $r(t)=\left \langle $\frac{1}{2}$t,\cos(t)\right \rangle$ in rectangular form by mapping the parametric coordinates to Cartesian coordinates :

$x=\frac{1}{2}$t$

We can now use this value to solve for :

$y=\sin(t)$

$y=\sin(2x)$

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Question

The graph of the vector function $r(t)=\left \langle $3t^{2}$-1,\cos(t)\right \rangle$ can also be represented by the graph of which of the following functions in rectangular form?

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Answer

We can find the graph of $r(t)=\left \langle $3t^{2}$-1,\cos(t)\right \rangle$ in rectangular form by mapping the parametric coordinates to Cartesian coordinates :

$t=$\sqrt{\frac{x+1}{3}$}$

We can now use this value to solve for :

$y=\cos$\sqrt{\frac{x+1}{3}$}$

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Question

What is the derivative of $r=4sin\theta+2$ ?

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Answer

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=4sin\theta+2$ , we must first find the derivative of with respect to $\theta$ as follows:

$$\frac{dr}{d\theta}$=4cos\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$$\frac{dy}{dx}$=\frac{\frac{dr}{d\theta}$sin\theta+rcos\theta}{$\frac{dr}{d\theta}$cos\theta-rsin\theta}$

$\frac{dy}{dx}$=\frac{4cos\theta\sin\theta+(4sin\theta+2)cos\theta}{4cos\theta\cos\theta-(4sin\theta+2)sin\theta}$

$\frac{dy}{dx}$=\frac{4cos\theta\sin\theta+4sin\theta\cos\theta+2\cos\theta}{4cos\theta\cos\theta-4sin\theta\sin\theta-2sin\theta}$

Using the trigonometric identity , we can deduce that . Swapping this into the denominator, we get:

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Question

What is the polar form of ?

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Answer

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given , then:

$r\sin\theta=\frac{3}$9{}(r\cos $\theta)^{2}$$

Dividing both sides by $r\cos\theta$ , we get:

$\tan \theta=\frac{3}$9{}r\cos\theta$

$$\frac{\tan \theta}{\cos\theta}$=\frac{3}{9}$r$

$\tan \theta\sec\theta=\frac{3}{9}$r$

$r=\frac{9}{3}$\tan \theta\sec\theta$

$r=3\tan \theta\sec\theta$

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Question

Rewrite in polar form:

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Answer

$\left (r \cos \theta \right ) ^{2} = 3\left (r \cos \theta \right )\left (r \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta = 3 $r^{2}$ \cos \theta; \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta - 3 $r^{2}$ \cos \theta; \sin \theta \right ) = 1$

$r ^{2} \left ( \ \cos ^{2} \theta - 3 \cos \theta; \sin \theta \right ) = 1$

$r ^{2} = $\frac{1}{\cos ^{2}$ \theta - 3 \cos \theta; \sin \theta }$

$r =$\sqrt{ \frac{1}{\cos ^{2}$ \theta - 3 \cos \theta; \sin \theta }}$

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Question

Graph the equation $r=cos(\theta )$ where $0\leq \theta \leq2\pi$ .

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Answer

At angle the graph as a radius of . As it approaches $\frac{\pi }{2}$ , the radius approaches .

As the graph approaches $\pi$ , the radius approaches .

Because this is a negative radius, the curve is drawn in the opposite quadrant between $\frac{3\pi }{2}$ and $2\pi$ .

Between $\pi$ and $\frac{3\pi }{2}$ , the radius approaches from and redraws the curve in the first quadrant.

Between $\frac{3\pi }{2}$ and $2\pi$ , the graph redraws the curve in the fourth quadrant as the radius approaches from .

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Question

Draw the graph of $r=cos(2\theta )$ from $0\leq \theta \leq2\pi$ .

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Answer

Because this function has a period of $\pi$ , the x-intercepts of the graph happen at a reference angle of $\frac{\pi }{4}$ (angles halfway between the angles of the axes).

Between and $\frac{\pi }{4}$ the radius approaches from .

Between $\frac{\pi }{4}$ and $\frac{\pi }{2}$ , the radius approaches from and is drawn in the opposite quadrant, the third quadrant because it has a negative radius.

From $\frac{\pi }{2}$ to $\frac{3\pi }{4}$ the radius approaches from , and is drawn in the fourth quadrant, the opposite quadrant.

Between $\frac{3\pi }{4}$ and $\pi$ , the radius approaches from .

From $\pi$ and $\frac{5\pi }{4}$ , the radius approaches from .

Between $\frac{5\pi }{4}$ and $\frac{3\pi }{2}$ , the radius approaches from . Because it is a negative radius, it is drawn in the opposite quadrant, the first quadrant.

Then between $\frac{3\pi }{2}$ and $\frac{7\pi }{4}$ the radius approaches from and is draw in the second quadrant.

Finally between $\frac{7\pi }{4}$ and $2\pi$ , the radius approaches from .

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Question

What is the following coordinate in polar form?

Provide the angle in degrees.

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Answer

To calculate the polar coordinate, use

$r=$\sqrt{29}$$

However, keep track of the angle here. 68 degree is the mathematical equivalent of the expression, but we know the point (-2,-5) is in the 3rd quadrant, so we have to add 180 to it to get 248.

Some calculators might already have provided you with the correct answer.

$\theta=248$ .

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Question

What is the equation in polar form?

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Answer

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ and $x=r\cos \theta$ . Given , then $r\sin \theta=2(r\cos \theta $)^{2}$$=2r^{2}$$\cos^{2}$ \theta$ .

$r\sin \theta=2(r\cos \theta $)^{2}$$=2r^{2}$$\cos^{2}$ \theta$ . Dividing both sides by $r\cos \theta$ ,

$\tan \theta=2r\cos \theta$

$$\frac{\tan \theta}{\cos \theta}$=2r$

$\tan \theta}{\sec \theta=2r$

$$\frac{1}{2}$\tan \theta}{\sec \theta=r$

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Question

What is the equation $y=\frac{6}{x}$$ in polar form?

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Answer

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ and $x=r\cos \theta$ . Given $y=\frac{6}{x}$$ , then $r\sin \theta=\frac{6}{r\cos \theta}$$ . Multiplying both sides by $r\cos \theta$ ,

$r=\sqrt$\frac{6}{\sin \theta cos \theta}$$

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Question

Convert the following function into polar form:

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Answer

The following formulas were used to convert the function from polar to Cartestian coordinates:

$x=r\cos(\theta), y=r\sin(\theta), $x^2$$+y^2$$=r^2$$

Note that the last formula is a manipulation of a trignometric identity.

Simply replace these with x and y in the original function.

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Question

What is the equation in polar form?

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Answer

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given , then:

$r\sin\theta=-$\frac{1}{2}$(r\cos $\theta)^{2}$$

Dividing both sides by $r\cos\theta$ , we get:

$\tan \theta=-$\frac{1}{2}$r\cos\theta$

$$\frac{\tan \theta}{\cos\theta}$=-$\frac{1}{2}$r$

$\tan \theta\sec\theta=-$\frac{1}{2}$r$

$r=-2\tan \theta\sec\theta$

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Question

What is the polar form of ?

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Answer

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given , then:

Dividing both sides by $r\cos\theta$ , we get:

$\tan\theta=-7r\cos\theta$

$$\frac{\tan\theta}{\cos\theta}$=-7r$

$\tan\theta\sec\theta=-7r$

$r=-$\frac{1}{7}$\tan\theta\sec\theta$

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Question

What is the polar form of ?

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Answer

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given , then:

Dividing both sides by $r\cos\theta$ , we get:

$\tan\theta=-r\cos\theta$

$$\frac{\tan\theta}{\cos\theta}$=-r$

$\tan\theta\sec\theta=-r$

$r=-\tan\theta\sec\theta$

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Question

What is the polar form of ?

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Answer

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given , then:

$r\sin\theta=-10(r\cos\theta)+5$

$r\sin\theta+10(r\cos\theta)=5$

$r(\sin\theta+10\cos\theta)=5$

$r=\frac{5}{\sin\theta+10\cos\theta}$$

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