Parametric, Polar, and Vector Functions - AP Calculus BC

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Rewrite as a Cartesian equation:

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

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$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}$$

Rewrite as a Cartesian equation:

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

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

, 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}}$ .

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

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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 .

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

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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.

Graph where $0\leq \theta \leq2\pi$ .

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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 $2\pi$ .

Draw the curve of from $0\leq \theta \leq2\pi$ .

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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 $2\pi$ .

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

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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}$=x-3$

Find the length of the following parametric curve

, , $0\leq t \leq 2 \pi$ .

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The length of a curve is found using the equation

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,

to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ .

In this case

,

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

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 $\sqrt{ab}$=$\sqrt{a}$$\sqrt{b}$

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

Using the trigonometric identity where is a constant and

$=\int_${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, , gives us the final solution

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

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

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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

![](https://lh5.googleusercontent.com/sZ_rj0xEI7Kq_Z-Edh2pS8Zc99IsjfAAUhXcDL_ljJitVl4frWT1jO0jGLRJ107h_MKStM2BhHRsKqWunfl6-EegWhTjqDQkFgHFn2y4zbd2mTLgkyWksjTY-PLvTjWJrh61PP4 $"$\frac{d}{dx}$x^{n}$$=nx^{n-1}$") for all $n\neq0$ , to derive

and

.

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}$$

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

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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

![](https://lh5.googleusercontent.com/sZ_rj0xEI7Kq_Z-Edh2pS8Zc99IsjfAAUhXcDL_ljJitVl4frWT1jO0jGLRJ107h_MKStM2BhHRsKqWunfl6-EegWhTjqDQkFgHFn2y4zbd2mTLgkyWksjTY-PLvTjWJrh61PP4 $"$\frac{d}{dx}$x^{n}$$=nx^{n-1}$") for all $n\neq0$ , to derive

and

.

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}$$

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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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:

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

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

Rewrite in polar form:

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

$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 }}$

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

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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 .

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

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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 .

What is the following coordinate in polar form?

Provide the angle in degrees.

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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.

$\theta=248$ .

What is the equation in polar form?

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

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

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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}$$

Convert the following function into polar form:

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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.

What is the equation in polar form?

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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}$r$

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

What is the polar form of ?

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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}$\tan\theta\sec\theta$