Parametric Form

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AP Calculus BC › Parametric Form

Questions 1 - 10

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

R_sin2x

R_cos2x

R_sinx

R_sinx_1

Faker_cosx

Explanation

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.

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

Explanation

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:

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

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

R2_sin2x

R2_cos2x

R_sinx

R_sinx_1

R_sin2x

Explanation

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 $2\pi$ .

Rewrite as a Cartesian equation:

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

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

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

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

Explanation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Explanation

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

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

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

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

R_sinx

R_cosx

Faker_cosx

R_sinx_1

R_sin2x

Explanation

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 .

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

R2_cos2x

R_cos2x

R_sin2x

R_cosx

R2_sin2x

Explanation

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 $2\pi$ .

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

$4\sqrt{5}$

$5\sqrt{5}$

$3\sqrt{5}$

$2\sqrt{5}$

$\sqrt{5}$

Explanation

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 , 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 ,

we can determine that:

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

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

$L=4\sqrt{5}$

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

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

$\frac{6}{5}$

$\frac{5}{6}$

$\frac{10}{3}$

$\frac{3}{10}$

$10\sqrt3$

Explanation

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.

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

$4\sqrt{13}$

$5\sqrt{13}$

$6\sqrt{13}$

$3\sqrt{13}$

$7\sqrt{13}$

Explanation

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 , 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 ,

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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