Parametric Form - AP Calculus BC

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Question

Convert the following parametric equation into rectangular form:

$x=2t, y=\cos(t)+2$

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Answer

To convert from parametric form to rectangular form, we must eliminate the parameter:

To do this we will first get t in terms of x.

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

Now, replace all of the t's in the equation for y:

$y=\cos\left(\frac{x}{2}\right)+2$

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Question

Convert the following equation from parametric to rectangular form:

$x=t^2, y=\sec(t)+\tan(t)$

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Answer

To convert from parametric to rectangular form, we must eliminate the parameter from one of the equations:

$t=\pm\sqrt{x}$

Now, replace all of the t's in the equation for y with the above term:

$y=\sec(\pm \sqrt{x})+\tan(\pm \sqrt{x})$

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Question

Convert the following parametric equation to rectangular form:

$x=2t, y=\ln(t)$

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Answer

To convert from parametric to rectangular coordinates, we must eliminate the parameter by finding t in terms of x or y:

We will start by taking the exponential of both sides of the equation . Recall that $e^{ln(t)}=t$ .

Therefore we get,

.

Now, replace t with the above term in the equation for x:

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Question

Convert the following parametric equation to rectangular form:

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Answer

To convert the equation into parametric form, we must eliminate the parameter by finding t in terms of either x or y:

Now, replace all of the t's in the equation for y with the above term:

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Question

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

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Answer

Given and , let's solve both equations for :

$x=6t-7\rightarrow t=\frac{x+7}{6}$

$y=8t+9\rightarrow t=\frac{y-9}{8}$

Since both equations equal , let's set them equal to each other and solve for :

$\frac{y-9}{8}=\frac{x+7}{6}$

$y-9=8\left(\frac{x+7}{6}\right)$

$y=\frac{4}{3}(x+7)+9$

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Question

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

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Answer

Given and , let's solve both equations for :

$x=9-t\rightarrowt=9-x$

$y=10-2t\rightarrow t=-\frac{y-10}{2}$

Since both equations equal , let's set them equal to each other and solve for :

$-\frac{y-10}{2}=9-x$

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Question

Given $x=-\frac{1}{2}t$ and , what is in terms of (rectangular form)?

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Answer

Given $x=-\frac{1}{2}t$ and , let's solve both equations for :

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

$y=3t\rightarrow t=\frac{y}{3}$

Since both equations equal , let's set them equal to each other and solve for :

$\frac{y}{3}=-2x$

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Question

When $x=-\frac{1}{3}t+8$ and $y=\frac{2}{3}t+7$ , what is in terms of (rectangular form)?

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Answer

Given $x=-\frac{1}{3}t+8$ and $y=\frac{2}{3}t+7$ , wet's solve both equations for :

$x=-\frac{1}{3}t+8\rightarrow t=-3(x-8)$

$y=\frac{2}{3}t+7\rightarrow t=\frac{3}{2}(y-7)$

Since both equations equal , let's set them equal to each other and solve for :

$\frac{3}{2}(y-7)=-3(x-8)$

$y-7=\frac{2}{3}(-3)(x-8)$

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Question

When $x=\frac{3}{4}t-5$ and , what is in terms of (rectangular form)?

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Answer

Given $x=\frac{3}{4}t-5$ and , wet's solve both equations for :

$x=\frac{3}{4}t-5\rightarrow t=\frac{4}{3}(x+5)$

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

Since both equations equal , let's set them equal to each other and solve for :

$y-4=(7)(\frac{4}{3})(x+5)$

$y=(7)(\frac{4}{3})(x+5)+4$

$y=\frac{28}{3}(x+5)+4$

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Question

Given $x=4+\frac{1}{2}t$ and , what is in terms of ?

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Answer

Given $x=4+\frac{1}{2}t$ and , wlet's solve both equations for :

$x=4+\frac{1}{2}t\rightarrow t=2(x-4)$

$y=-6t\rightarrow t=-\frac{1}{6}y$

Since both equations equal , let's set them equal to each other and solve for :

$-\frac{1}{6}y=2(x-4)$

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Question

When $x=-t+\frac{1}{2}$ and , what is in terms of (rectangular form)?

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Answer

Given $x=-t+\frac{1}{2}$ and , wet's solve both equations for :

$x=-t+\frac{1}{2}\rightarrow t=\frac{1}{2}-x$

$y=3t-2\rightarrow t=\frac{y+2}{3}$

Since both equations equal , let's set them equal to each other and solve for :

$\frac{y+2}{3}=\frac{1}{2}-x$

$y+2=(3)(\frac{1}{2}-x)$

$y=(3)(\frac{1}{2}-x)-2$

$y=\frac{3}{2}-3x-2$

$y=\frac{3}{2}-3x-\frac{4}{2}$

$y=-3x-\frac{1}{2}$

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Question

Given and , what is in terms of ?

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Answer

Given and , let's solve both equations for :

$x=7-t\rightarrow t=7-x$

$y=2t+5\rightarrow t=\frac{y-5}{2}$

Since both equations equal , let's set them equal to each other and solve for :

$\frac{y-5}{2}=7-x$

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Question

Write in Cartesian form:

$x = 3t, y = 2t^{2}$

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Answer

, so

$x^{2} = (3t)^{2} = 9t^{2} = \frac{9}{2} \cdot 2t^{2} = \frac{9}{2} y$ .

$\frac{9}{2} y= x^{2}$ , so

$\frac{2}{9}\cdot \frac{9}{2} y = \frac{2}{9}x^{2}$

$y = \frac{2}{9}x^{2}$

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Question

Write in Cartesian form:

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Answer

,

so the Cartesian equation is

.

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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

Write in Cartesian form:

$x = \ln (t+1)$

$t \in (-1, \infty )$

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Answer

$x = \ln (t+1)$

so

$e ^{x} = \left [ \ln (t+1) \right ] ^{x}$

$e ^{x} = t + 1$

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

Therefore the Cartesian equation is $y = e^{2x} + 1$ .

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Question

Write in Cartesian form:

$x = \cos 2t, y = \sin t$

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Answer

Rewrite using the double-angle formula:

$x = \cos 2t =1 - 2\sin ^{2} t = 1 - 2y^{2}$

Then

$x = 1 - 2y^{2}$

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

which is the correct choice.

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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

$\small x=3t+4$ and $\small y=1/t$ . What is $\small y$ in terms of $\small x$ (rectangular form)?

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Answer

In order to solve this, we must isolate $\small t$ in both equations.

$\small \small x=3t+4\rightarrow t=(x-4)/3$ and

$\small y=1/t\rightarrow t=1/y$ .

Now we can set the right side of those two equations equal to each other since they both equal $\small t$ .

$\small (x-4)/3=1/y$ .

By multiplying both sides by $\small 3y/(x-4)$ , we get $\small y=3/(x-4)$ , which is our equation in rectangular form.

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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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