AP Calculus BC - Parametric Form

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

$y=\frac{2}{5}(x+9)+7$

None of the answers provided

$y=\frac{2}{5}(x-9)+7$

$y=\frac{2}{5}(x-9)-7$

$y=\frac{2}{5}(x-9)+7$

Explanation

Given and , let's solve both equations for :

$x=5t-9\rightarrow t=\frac{x+9}{5}$

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

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

$\frac{y-7}{2}=\frac{x+9}{5}$

$y-7=2\left(\frac{x+9}{5}\right)$

$y=\frac{2}{5}(x+9)+7$

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

$y=\frac{5}{9}({x+2}})+1$

$y=\frac{2}{9}({x+5}})+1$

$y=\frac{1}{9}({x+2}})+5$

$y=\frac{9}{5}({x+2}})+1$

$y=\frac{9}{2}({x+5}})+1$

Explanation

Knowing that and , we can isolate in both equations as follows:

$x=9t-2\rightarrow t=\frac{x+2}{9}$

$y=5t+1\rightarrow t=\frac{y-1}{5}$

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

$\frac{y-1}{5}=\frac{x+2}{9}$

${y-1}=5(\frac{x+2}{9})$

$y=\frac{5}{9}({x+2}})+1$

Given $x=\frac{1}{2}t-2$ and $y=\frac{1}{3}t-5$ , what is in terms of (rectangular form)?

$y=\frac{2}{3}(x+2)-5$

$y=\frac{2}{3}(x+2)+5$

$y=\frac{2}{3}(x-2)-5$

$y=\frac{2}{3}(x-2)+5$

$y=-\frac{2}{3}(x+2)-5$

Explanation

Given $x=\frac{1}{2}t-2$ and $y=\frac{1}{3}t-5$ , let's solve both equations for :

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

$y=\frac{1}{3}t-5\rightarrow t=3(y+5)$

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

$y+5=\frac{2}{3}(x+2)$

$y=\frac{2}{3}(x+2)-5$

Convert the following to rectangular form from parametric form:

$y=\sqrt[3]{\left(\frac{x-5}{2}\right)}$

$y=\left(\frac{x-5}{2}\right)^{\frac{3}{2}}$

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

Explanation

To convert a parametric equation into a rectangular equation, we must eliminate the parameter. We were already given an equation where t was in terms of just a variable:

Next, substitute this into the equation for x which contains t:

Finally, rearrage and solve for y:

$y=\sqrt[3]{\left(\frac{x-5}{2}\right)}$

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

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

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

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

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

Explanation

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.

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

$y=9-\frac{x+11}{2}$

$y=9+\frac{x+11}{2}$

$y=9-\frac{x-11}{2}$

$y=9+\frac{x-11}{2}$

None of the above

Explanation

Given and , let's solve both equations for :

$x=2t-11\rightarrow t=\frac{x+11}{2}$

$y=-t+9\rightarrow t=9-y$

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

$9-y=\frac{x+11}{2}$

$y=9-\frac{x+11}{2}$

Convert the following parametric equation to rectangular form:

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

Explanation

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:

If and , what is in terms of ?

$y=\frac{5}{3}(x-14)+12$

$y=\frac{5}{3}(x+14)+12$

$y=\frac{5}{3}(x+14)-12$

$y=\frac{5}{3}(x-14)-12$

None of the above

Explanation

Given and , let's solve both equations for :

$x=3t+14\rightarrow t=\frac{x-14}{3}$

$y=5t+12\rightarrow t=\frac{y-12}{5}$

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

$\frac{y-12}{5}=\frac{x-14}{3}$

$y-12=\frac{5}{3}(x-14)$

$y=\frac{5}{3}(x-14)+12$

Find the arc length of the curve:

$x=8\cos(3t),,y=8\sin(3t),,0\leq t\leq \frac{\pi}{2}$

$12\pi$

$4\pi$

$8\pi$

$24\pi$

$\frac{3\pi}{2}$

Explanation

Finding the length of the curve requires simply applying the formula:

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

Where:

$a=0, b =\frac{\pi}{2}$

Since we are also given and , we can easily compute the derivatives of each:

$\frac{dx}{dt}=\frac{d}{dt}[8\cos(3t)]=-8\sin(3t)\cdot3=-24\sin(3t)$

$\frac{dy}{dt}=\frac{d}{dt}[8\sin(3t)]=8\cos(3t)\cdot3=24\cos(3t)$

Applying these into the above formula results in:

$L=\int_{0}^{\frac{\pi}{2}}\sqrt{(-24\sin(3t))^2+(24\cos(3t))^2},dt=\int_{0}^{\frac{\pi}{2}}\sqrt{24^2\sin^2(3t)+24^2\cos^2(3t)},dt$ We can factor out the common , and pull it outside of the square-root, and we will notice one of the most common trigonometric identities: