AP Calculus BC : Parametric, Polar, and Vector Functions

Study concepts, example questions & explanations for AP Calculus BC

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Example Questions

Example Question #1 : Parametric, Polar, And Vector

Rewrite as a Cartesian equation:

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

Possible Answers:

\(\displaystyle y = x + 4 + 4\sqrt{x}\)

\(\displaystyle y = x - 2\)

\(\displaystyle y = x - 4 - 4\sqrt{x}\)

\(\displaystyle y = x+2\)

\(\displaystyle y = x + 4 - 4\sqrt{x}\)

Correct answer:

\(\displaystyle y = x + 4 - 4\sqrt{x}\)

Explanation:

\(\displaystyle x = t^{2} + 2t + 1\)

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

\(\displaystyle \pm \sqrt{x} = t + 1\)

So 

\(\displaystyle \sqrt{x} = t + 1\) or \(\displaystyle -\sqrt{x} = t + 1\)

We are restricting \(\displaystyle t\) to values on \(\displaystyle \left [ -1, 1\right ]\), so \(\displaystyle t + 1\) is nonnegative; we choose 

\(\displaystyle \sqrt{x} = t + 1\).

Also,

\(\displaystyle y = t^{2} - 2t + 1\)

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

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

So 

\(\displaystyle \sqrt{y} = t- 1\) or \(\displaystyle -\sqrt{y} = t- 1\)

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

\(\displaystyle -\sqrt{y} = t- 1\)

or equivalently,

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

to make \(\displaystyle t - 1\) nonpositive.

 

Then,

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

and 

\(\displaystyle \sqrt{x}+ \sqrt{y} = 2\)

\(\displaystyle \sqrt{y} = 2 - \sqrt{x}\)

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

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

\(\displaystyle y = x + 4 - 4\sqrt{x}\)

Example Question #2 : Parametric, Polar, And Vector Functions

Rewrite as a Cartesian equation:

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

Possible Answers:

\(\displaystyle y = \frac{ 2x ^{2 \log e} -1}{2 x ^{\log e}}\)

\(\displaystyle y = \frac{ x ^{2 \log e} +1}{2 x ^{\log e}}\)

\(\displaystyle y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}\)

\(\displaystyle y = \frac{ x ^{2 \log e} }{2 x ^{\log e}}\)

\(\displaystyle y = \frac{ x ^{2 \log e} -1}{ x ^{2 \log e} + 1}\)

Correct answer:

\(\displaystyle y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}\)

Explanation:

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

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

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

This makes the Cartesian equation

\(\displaystyle y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}\).

Example Question #3 : Parametric, Polar, And Vector Functions

If \(\displaystyle x=3+t\) and \(\displaystyle y=4t+7\), what is \(\displaystyle y\) in terms of \(\displaystyle x\) (rectangular form)?

Possible Answers:

\(\displaystyle y=4x-7\)

\(\displaystyle y=4x-2\)

\(\displaystyle y=4x-12\)

\(\displaystyle y=4x-5\)

\(\displaystyle y=4x-3\)

Correct answer:

\(\displaystyle y=4x-5\)

Explanation:

Given \(\displaystyle x=3+t\) and  \(\displaystyle y=4t+7\), we can find \(\displaystyle y\) in terms of \(\displaystyle x\) by isolating \(\displaystyle t\) in both equations:

 

\(\displaystyle x=3+t\rightarrow t=x-3\)

\(\displaystyle y=4t+7\rightarrow t=\frac{y-7}{4}\)

Since both of these transformations equal \(\displaystyle t\), we can set them equal to each other:

\(\displaystyle \frac{y-7}{4}=x-3\)

\(\displaystyle y-7=4(x-3)\)

\(\displaystyle y-7=4x-12\)

\(\displaystyle y=4x-5\)

Example Question #191 : Ap Calculus Bc

Given \(\displaystyle x=t+10\) and \(\displaystyle y=2t-9\), what is the arc length between \(\displaystyle 0\leqt\leq t\leq4\)?

Possible Answers:

\(\displaystyle 2\sqrt{5}\)

\(\displaystyle 4\sqrt{5}\)

\(\displaystyle \sqrt{5}\)

\(\displaystyle 3\sqrt{5}\)

\(\displaystyle 5\sqrt{5}\)

Correct answer:

\(\displaystyle 4\sqrt{5}\)

Explanation:

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

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

Given  \(\displaystyle x=t+10\) and \(\displaystyle y=2t-9\), we can use using the Power Rule

 for all , to derive 

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

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

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

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

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

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

Now, using the Power Rule for Integrals

 for all ,

we can determine that:

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

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

\(\displaystyle L=4\sqrt{5}\)

Example Question #4 : Parametric, Polar, And Vector Functions

Given \(\displaystyle x=4t-5\) and \(\displaystyle y=6t+2\), what is the length of the arc from \(\displaystyle 0\leq t\leq2\)?

Possible Answers:

\(\displaystyle 3\sqrt{13}\)

\(\displaystyle 4\sqrt{13}\)

\(\displaystyle 5\sqrt{13}\)

\(\displaystyle 6\sqrt{13}\)

\(\displaystyle 7\sqrt{13}\)

Correct answer:

\(\displaystyle 4\sqrt{13}\)

Explanation:

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

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

Given \(\displaystyle x=4t-5\) and \(\displaystyle y=6t+2\), we can use using the Power Rule

for all  , to derive

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

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

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

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

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

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

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

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

Now, using the Power Rule for Integrals

for all ,

we can determine that:

\(\displaystyle L=[2t\sqrt{13}]_{0}^{2}\textrm{}\)

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

\(\displaystyle L=4\sqrt{13}\)

Example Question #151 : Parametric, Polar, And Vector

Find the length of the following parametric curve 

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

Possible Answers:

\(\displaystyle sin(t)+cos(t)\)

\(\displaystyle e^{2 \pi}\)

\(\displaystyle e^{t}cos(t)\)

\(\displaystyle \sqrt{2}\left ( e^{2 \pi}-1}\right )\)

\(\displaystyle 1\)

Correct answer:

\(\displaystyle \sqrt{2}\left ( e^{2 \pi}-1}\right )\)

Explanation:

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

We use the product rule,

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

the trigonometric rule,

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

and exponential rule,

\(\displaystyle \frac{d}{dt}[e^{t}]=e^{t}\) to find \(\displaystyle \frac{dx}{dt}\) and \(\displaystyle \frac{dy}{dt}\)

In this case

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

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

\(\displaystyle =e^{t}*(-sin(t))+cos(t)*e^{t}=e^{t}cos(t)-e^{t}sin(t)\)

 

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

\(\displaystyle =e^{t}cos(t)+sin(t)*e^{t}=e^{t}cos(t)+e^{t}sin(t)\)

\(\displaystyle \alpha=0, \beta=2\pi\)

The length of this curve is

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

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

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

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

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

Using the identity \(\displaystyle \sqrt{a*b}=\sqrt{a}\sqrt{b}\)

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

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

\(\displaystyle =\int_{0}^{2\pi }\sqrt{2}e^{t}dt}=\sqrt{2}\int_{0}^{2\pi }e^{t}dt}\)

Using the exponential rule, \(\displaystyle \int e^{t}dt}=e^{t}\)

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

Using the exponential rule, \(\displaystyle e^{0}=1\), gives us the final solution

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

 

 

Example Question #4 : Parametric, Polar, And Vector Functions

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

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

Possible Answers:

\(\displaystyle \frac{3}{10}\)

\(\displaystyle \frac{6}{5}\)

\(\displaystyle \frac{5}{6}\)

\(\displaystyle \frac{10}{3}\)

\(\displaystyle 10\sqrt3\)

Correct answer:

\(\displaystyle \frac{6}{5}\)

Explanation:

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

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

From the problem statement:

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

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

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

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

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

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

This is one of the answer choices.

Example Question #7 : Parametric, Polar, And Vector Functions

Draw the graph of \(\displaystyle r=sin\theta\) from \(\displaystyle 0\leq \theta \leq2\pi\).

Possible Answers:

R_sinx

Faker_cosx

R_sinx_1

R_sin2x

R_cosx

Correct answer:

R_sinx

Explanation:

Between \(\displaystyle 0\) and \(\displaystyle \frac{\pi }{2}\), the radius approaches \(\displaystyle 1\) from \(\displaystyle 0\).

From \(\displaystyle \frac{\pi }{2}\) to \(\displaystyle \pi\) the radius goes from \(\displaystyle 1\) to \(\displaystyle 0\).

Between \(\displaystyle \pi\) and \(\displaystyle \frac{3\pi }{2}\), the curve is redrawn in the opposite quadrant, the first quadrant as the radius approaches \(\displaystyle -1\).

From \(\displaystyle \frac{3\pi }{2}\) and \(\displaystyle 2\pi\), the curve is redrawn in the second quadrant as the radius approaches \(\displaystyle 0\) from \(\displaystyle -1\).   

Example Question #1 : Parametric, Polar, And Vector Functions

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

Possible Answers:

R_cos2x

R_sin2x

R_sinx_1

Faker_cosx

R_sinx

Correct answer:

R_sin2x

Explanation:

Because this function has a period of \(\displaystyle \pi\), the amplitude of the graph \(\displaystyle y=sin(2x)\)  appear at a reference angle of \(\displaystyle \frac{\pi }{4}\) (angles halfway between the angles of the axes).  

Between \(\displaystyle 0\) and \(\displaystyle \frac{\pi }{4}\) the radius approaches 1 from 0.

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

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

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

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

Then between \(\displaystyle \frac{3\pi }{2}\) and \(\displaystyle \frac{7\pi }{4}\) 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 \(\displaystyle \frac{7\pi }{4}\) and \(\displaystyle 2\pi\), the curve is drawn in the second quadrant.                  

Example Question #5 : Parametric, Polar, And Vector Functions

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

Possible Answers:

R_cos2x

R_sin2x

R2_cos2x

R_cosx

R2_sin2x

Correct answer:

R2_cos2x

Explanation:

Taking the graph of \(\displaystyle y=cos(2x)\), 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 \(\displaystyle 0\) to \(\displaystyle \frac{\pi }{4}\)\(\displaystyle \frac{3\pi }{4}\) to \(\displaystyle \frac{5\pi }{4}\), and \(\displaystyle \frac{7\pi }{4}\) to \(\displaystyle 2\pi\).

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 \(\displaystyle \pm 1\).

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

From \(\displaystyle \frac{3\pi }{4}\) to \(\displaystyle \pi\), 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 \(\displaystyle \pi\) to \(\displaystyle 2\pi\).    

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