Parametric, Polar, and Vector Functions - AP Calculus BC

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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:

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

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

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

$\vec{v}+\vec{u}=(x^2 + 2,2x,16)$

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Question

Calculate the polar form hypotenuse of the following cartesian equation:

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Answer

In a cartesian form, the primary parameters are and . In polar form, they are and $\theta$

is the hypotenuse, and $\theta$ is the angle created by .

2 things to know when converting from Cartesian to polar.

$x= r\cos(\theta)$

$y= r\sin(\theta)$

You want to calculate the hypotenuse,

Solution:

$y=10x^4=r\sin(\theta)$

$y=10(r\cos(\theta))^4$

$y=10r^4\cos^4(\theta) = r\sin(\theta)$

$r^3 = \frac{\tan(\theta)}{10\cos^3(\theta)}$

$r = \sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

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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

Rewrite as a Cartesian equation:

$x = t^{2} + 2t + 1, y = t^{2} - 2t + 1, t \in [-1, 1]$

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Answer

$x = t^{2} + 2t + 1$

$x = \left ( t + 1\right )^{2}$

$\pm \sqrt{x} = t + 1$

So

$\sqrt{x} = t + 1$ or $-\sqrt{x} = t + 1$

We are restricting to values on $\left [ -1, 1\right ]$ , so is nonnegative; we choose

$\sqrt{x} = t + 1$ .

Also,

$y = t^{2} - 2t + 1$

$y= \left ( t - 1\right )^{2}$

$\sqrt{y} = \pm \left ( t - 1\right )$

So

$\sqrt{y} = t- 1$ or $-\sqrt{y} = t- 1$

We are restricting to values on $\left [ -1, 1\right ]$ , so is nonpositive; we choose

$-\sqrt{y} = t- 1$

or equivalently,

$\sqrt{y} = -t+ 1$

to make nonpositive.

Then,

$\sqrt{x}+ \sqrt{y} = (t + 1) + (-t + 1)$

and

$\sqrt{x}+ \sqrt{y} = 2$

$\sqrt{y} = 2 - \sqrt{x}$

$y =\left ( 2 - \sqrt{x} \right )^2$

$y = 4 - 4\sqrt{x} + \left (\sqrt{x} \right )^{2}$

$y = x + 4 - 4\sqrt{x}$

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Question

Rewrite as a Cartesian equation:

$x =10^ {t}, y = \sinh t, t \in (0,\infty )$

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Answer

$y = \sinh t = \frac{e ^{2t} -1}{2e ^{t}} = \frac{\left (e ^{ t} \right) ^{2} -1}{2e ^{t}}$

$e^{t} = 10 ^{\log e \cdot t} =\left ( 10 ^{ t} \right )^{\log e} = x ^{\log e}$ , so

$y = \frac{\left ( x ^{\log e} \right) ^{2} -1}{2 x ^{\log e}}= \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$

This makes the Cartesian equation

$y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$ .

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Question

Draw the graph of $r=sin\theta$ from $0\leq \theta \leq2\pi$ .

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Answer

Between and $\frac{\pi }{2}$ , the radius approaches from .

From $\frac{\pi }{2}$ to $\pi$ the radius goes from to .

Between $\pi$ and $\frac{3\pi }{2}$ , the curve is redrawn in the opposite quadrant, the first quadrant as the radius approaches .

From $\frac{3\pi }{2}$ and $2\pi$ , the curve is redrawn in the second quadrant as the radius approaches from .

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Question

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

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Answer

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

Between and $\frac{\pi }{4}$ the radius approaches 1 from 0.

Between $\frac{\pi }{4}$ and $\frac{\pi }{2}$ , the radius approaches 0 from 1.

From $\frac{\pi }{2}$ to $\frac{3\pi }{4}$ the radius approaches -1 from 0 and is drawn in the opposite quadrant, the fourth quadrant because it has a negative radius.

Between $\frac{3\pi }{4}$ and $\pi$ , the radius approaches 0 from -1, and is also drawn in the fourth quadrant.

From $\pi$ and $\frac{5\pi }{4}$ , the radius approaches 1 from 0. Between $\frac{5\pi }{4}$ and $\frac{3\pi }{2}$ , the radius approaches 0 from 1.

Then between $\frac{3\pi }{2}$ and $\frac{7\pi }{4}$ the radius approaches -1 from 0. Because it is a negative radius, it is drawn in the opposite quadrant, the second quadrant. Likewise, as the radius approaches 0 from -1. Between $\frac{7\pi }{4}$ and $2\pi$ , the curve is drawn in the second quadrant.

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Question

Graph $r^2=cos(2\theta )$ where $0\leq \theta \leq2\pi$ .

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Answer

Taking the graph of , we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from to $\frac{\pi }{4}$ , $\frac{3\pi }{4}$ to $\frac{5\pi }{4}$ , and $\frac{7\pi }{4}$ to $2\pi$ .

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of $\pm 1$ .

To draw the graph, the radius is 1 at and traces to 0 at $\frac{\pi }{4}$ . As well, the negative part of the radius starts at -1 and traces to zero in the opposite quadrant, the third quadrant.

From $\frac{3\pi }{4}$ to $\pi$ , the curves are traced from 0 to 1 and 0 to -1 in the fourth quadrant. Following this pattern, the graph is redrawn again from the areas included in $\pi$ to $2\pi$ .

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Question

Draw the curve of $r^2=sin(2\theta )$ from $0\leq \theta \leq2\pi$ .

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Answer

Taking the graph of , we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from to $\frac{\pi }{2}$ and $\pi$ to $\frac{3\pi }{2}$ .

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of $\pm 1$ .

To draw the graph, the radius is 0 at and traces to 1 at $\frac{\pi }{4}$ . As well, the negative part of the radius starts at 0 and traces to-1 in the opposite quadrant, the third quadrant.

From $\frac{\pi }4{}$ to $\frac{\pi }{2}$ , the curves are traced from 1 to 0 and -1 to 0 in the third quadrant.

Following this pattern, the graph is redrawn again from the areas included in $\pi$ to $2\pi$ .

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Question

If and , what is in terms of (rectangular form)?

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Answer

Given and , we can find in terms of by isolating in both equations:

$x=3+t\rightarrow t=x-3$

$y=4t+7\rightarrow t=\frac{y-7}{4}$

Since both of these transformations equal , we can set them equal to each other:

$\frac{y-7}{4}=x-3$

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Question

Find the length of the following parametric curve

$x=e^{t}cos(t)$ , $y=e^{t}sin(t)$ , $0\leq t \leq 2 \pi$ .

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Answer

The length of a curve is found using the equation $L=\int_{\alpha }^{\beta }\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}dt$

We use the product rule,

$\frac{d}{dt}(ab)=a\frac{db}{dt}+b\frac{da}{dt}$ , when and are functions of ,

the trigonometric rule,

$\frac{d}{dt}[cos(t)]=-sin(t)$ and $\frac{d}{dt}[sin(t)]=cos(t)$

and exponential rule,

$\frac{d}{dt}[e^{t}]=e^{t}$ to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ .

In this case

$x=e^{t}cos(t)$ , $y=e^{t}sin(t)$

$\frac{dx}{dt}=\frac{d}{dt}e^{t}cos(t)=e^{t}\frac{d}{dt}cos(t)+cos(t)\frac{d}{dt}e^{t}$

$=e^{t}(-sin(t))+cos(t)e^{t}=e^{t}cos(t)-e^{t}sin(t)$

$\frac{dy}{dt}=\frac{d}{dt}e^{t}sin(t)=e^{t}\frac{d}{dt}sin(t)+sin(t)\frac{d}{dt}e^{t}$

$=e^{t}cos(t)+sin(t)e^{t}=e^{t}cos(t)+e^{t}sin(t)$

$\alpha=0, \beta=2\pi$

The length of this curve is

$L=\int_{0}^{2\pi }\sqrt{\left (e^{t}cos(t)-e^{t}sin(t)\right )^{2}+\left (e^{t}cos(t)+e^{t}sin(t)\right )^{2}}dt$

Using the identity $(a\pm b)^{2}=a^{2}\pm2ab+b^{2}$

$L=\int_{0}^{2\pi }\sqrt{\binom{\left (e^{2t}cos^{2}(t)-2e^{t}cos(t)e^{t}sin(t)+e^{2t}sin^{2}(t)\right )}{\left +(e^{2t}cos^{2}(t)+2e^{t}cos(t)e^{t}sin(t)+e^{2t}sin^{2}(t)\right)}dt}$

$=\int_{0}^{2\pi }\sqrt{(2e^{2t}cos^{2}(t)+2e^{2t}sin^{2}(t))dt}$

$=\int_{0}^{2\pi }\sqrt{2e^{2t}(cos^{2}(t)+sin^{2}(t))dt}$

Using the identity $\sqrt{a*b}=\sqrt{a}\sqrt{b}$

$L=\int_{0}^{2\pi }\sqrt{2e^{2t}} \sqrt{cos^{2}(t)+sin^{2}(t)dt}$

Using the trigonometric identity $sin^{2}(at)+cos^{2}(at)=1$ where is a constant and $e^{2t}=(e^{t})^2$

$=\int_{0}^{2\pi }\sqrt{2}e^{t}dt}=\sqrt{2}\int_{0}^{2\pi }e^{t}dt}$

Using the exponential rule, $\int e^{t}dt}=e^{t}$

$\sqrt{2}\int_{0}^{2\pi }e^{t}dt}=\sqrt{2}e^{t}|_{0}^{2\pi}=\sqrt{2}\left ( e^{2 \pi}-e^{0}}\right )$

Using the exponential rule, $e^{0}=1$ , gives us the final solution

$L=\sqrt{2}\left ( e^{2 \pi}-1}\right )$

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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

Given and , what is the arc length between $0\leqt\leq t\leq4$ ?

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Answer

In order to find the arc length, we must use the arc length formula for parametric curves:

$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$ .

Given and , we can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , to derive

$\frac{dx}{dt}=(1)t^{1-1}+(0)10=1$ and

$\frac{dy}{dt}=(1)2t^{1-1}-(0)9=2$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{0}^{4}\sqrt{(1)^{2}+(2)^{2}}dt$

$L=\int_{0}^{4}(\sqrt{1+4})dt$

$L=\int_{0}^{4}(\sqrt{5})dt$

Now, using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$L=[t\sqrt{5}]_{0}^{4}\textrm{}dt$

$L=(4)\sqrt{5}-(0)\sqrt{5}$

$L=4\sqrt{5}$

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Question

Given and , what is the length of the arc from $0\leq t\leq2$ ?

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Answer

In order to find the arc length, we must use the arc length formula for parametric curves:

$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$ .

Given and , we can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , to derive

$\frac{dx}{dt}=(1)4t^{1-1}-(0)5=4$ and

$\frac{dy}{dt}=(1)6t^{1-1}+(0)2=6$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{0}^{2}(\sqrt{(4)^{2}+(6)^{2}}dt$

$L=\int_{0}^{2}(\sqrt{16+36})dt$

$L=\int_{0}^{2}(\sqrt{52})dt$

$L=\int_{0}^{2}(\sqrt{4\times13})dt$

$L=\int_{0}^{2}(2\sqrt{13})dt$

Now, using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$L=[2t\sqrt{13}]_{0}^{2}\textrm{}$

$L=2(2)\sqrt{13}-2(0)\sqrt{13}$

$L=4\sqrt{13}$

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Question

Find dy/dx at the point corresponding to the given value of the parameter without eliminating the parameter:

$x=5\sec(t),y=3\tan(t), t=\frac{\pi}{6}$

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Answer

The formula for dy/dx for parametric equations is given as:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

From the problem statement:

$\frac{dy}{dt}=3\sec^2(t),\frac{dx}{dt}=5\sec(t)\tan(t)$

If we plug these into the above equation we end up with:

$\frac{dy}{dx}=\frac{3\sec^2(t)}{5\sec(t)\tan(t)}=\frac{3\sec(t)}{5\tan(t)}=\frac{3\frac{1}{cos(t)}}{5\frac{sin(t)}{cos(t)}}$

$\frac{dy}{dx}=\frac{3}{5}\frac{1}{\cos(t)}\frac{\cos(t)}{\sin(t)}=\frac{3}{5}\frac{1}{\sin(t)}$

If we plug in our given value for t, we end up with:

$\frac{3}{5}\frac{1}{\sin(\frac{\pi}{6})}=\frac{3}{5}\frac{1}{\frac{1}{2}}=\frac{3}{5}*2=\frac{6}{5}$

This is one of the answer choices.

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Question

Rewrite in polar form:

$x^{2} = 3xy + 1$

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Answer

$x^{2} = 3xy + 1$

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

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

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

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

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

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

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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

$r=\sqrt{x^2+y^2}$

$r=\sqrt{29}$

$\theta=tan^{-1}\left(\frac{y}{x}\right)=68$

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 $y=2x^{2}$ 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 $y=2x^{2}$ , 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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