Parametric Curves - AP Calculus BC

Card 0 of 40

Question

Find the length of the parametric curve described by

$x = 2\sin t$

$y = 2\cos t$

from to $\pi$ .

Answer

There are several ways to solve this problem, but the most effective would be to notice that we can derive the following-

$x^2 = 4\sin^2t$

$y^2 = 4\cos^2t$

Hence

$x^2+y^2 = 4\sin^2t + 4\cos^2t = 4$

Therefore our curve is a circle of radius $\sqrt4=2$ , and it's circumfrence is $4\pi$ . But we are only interested in half that circumfrence ( is from to $\pi$ , not $2\pi$ .), so our answer is $2\pi$ .

Alternatively, we could've found the length using the formula

$l = \int_{t_0}^{t_1}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$ .

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Question

Find the coordinates of the curve function

when .

Answer

To find the coordinates, we set into the curve function.

We get

and thus

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Question

Find the coordinates of the curve function

when $t=\frac{\pi}{2}$

Answer

To find the coordinates, we evaluate the curve function for $t=\pi/2$

As such,

$c(\frac{\pi}{2})=<cos (\frac{\pi}{2}), sin (\frac{\pi}{2}), cos (\frac{\pi}{2})>=<0,1,0>$

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Question

Find the coordinates of the curve function

when

Answer

To find the coordinates, we evaluate the curve function for

As such,

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Question

Find the equation of the line passing through the two points, given in parametric form:

Answer

To find the equation of the line passing through these two points, we must first find the vector between them:

$\vec{v}=\left \langle 1, 6, 2\right \rangle$

This was done by finding the difference between the x, y, and z components for the vectors. (This can be done in either order, it doesn't matter.)

Now, pick a point to be used in the equation of the line, as the initial point. We write the equation of line as follows:

$\vec{r}=\left \langle 1, 2, 3\right \rangle + t\left \langle 1, 6, 2\right \rangle=\left \langle 1+t, 2+6t, 3+2t\right \rangle$

The choice of initial point is arbitrary.

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Question

Find the coordinate of the parametric curve when

$t=\pi/2$

,

Answer

To find the coordinates of the parametric curve we plug in for

$t=\pi/2$ .

$x=sin,(\pi/2)= 1$

$y=(\pi/2)^2=\pi^2/4$

As such the coordinates are

$(1,\pi^2/4)$

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Question

Write the parametric equations of the line that passes through the points $\left ( 0,1,3\right )$ and $\left ( 2,5,8\right )$ .

Answer

First, you must find the vector that is parallel to the line.

This vector is

$v=\left \langle x_1-x_2,y_1-y_2,z_1-z_2\right \rangle$ .

From the points we were given, this becomes

$\left \langle -2,-4,-5\right \rangle$ .

To form the parametric equations, we need to pick a point that lies on the line we want.

The point $\left ( 0,1,3\right )$ is used.

The vector form of the line is from the following equation

$r=\left ( x_1,y_1,z_1\right )+t\left \langle v\right \rangle=\left ( 0,1,3\right )+t\left \langle -2,-4,-5\right \rangle$ .

We then rewrite each expression in terms of the variables x, y, and z.

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Question

Evaluate the line integral $\int_C f(x,y)ds$ of the function

over the line segment from to

Answer

Evaluate the line integral $\int_C f(x,y)ds$ using the function

$f(x,y)=y+\cos(\pi x)$ over the line segment from to

Define the Parametric Equations to Represent

The points given lie on the line $y = \frac{5}{4}x$ . Define the parameter , then can be written $y =\frac{5}{4}t$ . Therefore, the parametric equations for are:

$y(t)=\frac{5}{4}t$

_________________________________________________________________

The line integral of a function along the curve with the parametric equation and with $a\leq t\leq b$ is defined by:

$\int_C f(x,y)ds = \int_a^bf(x(t),y(t))\left | \vec{r'}(t)\right |dt$ (1)

Where $\vec{r'(t)}$ is the vector derivative of the vector $\vec{r}(t)=\left<x(t),y(t)\right>$ , therefore $\left | \vec{r'(t)}\right |$ is simply the magnitude of the vector derivative.

______________________________________________________________

Write the vector $\vec{r(t)}$ :

$\vec{r(t)}=\left<t, \frac{5}{4}t\right>$

Differentiate,

$\vec{r'(t)}=\left<1,\frac{5}{4}\right>$

The absolute value (magnitude) of this vector is:

$\left | \vec{r'}(t)\right |=\left | \left<1,\frac{5}{4 }\right>\right |=\sqrt{1^2 +\left(\frac{5}{4} \right )^2}=\frac{\sqrt{41}}{4}$

Write the function in terms of the parameter :

$f(x(t),y(t))=f\left(t,\frac{5}{4}t\right) =\frac{5}{4}t+\cos(\pi t)$

Insert everything into Equation (1) noting that the limits of integration will be $0\leq t \leq 4$ due to the fact that the parameter varies from to over the line segment we are integrating over.

$\int_C(y+\cos(\pi x))ds=\int_0^4\left(\frac{5}{4}t+\cos(\pi t)\right)\frac{\sqrt{41}}{4}dt$

$=\frac{\sqrt{41}}{4}\left[\frac{5}{8}t^2+\frac{1}{\pi}\sin(\pi t) \right ]_{t=0}^{t=4}$

$=\frac{\sqrt{41}}{4}\left[\frac{5}{8}(16)+\frac{1}{\pi }\sin(4 \pi) -0-\frac{1}{\pi}\sin(0)\right ]$

$=\frac{5\sqrt{41}}{2}$

Compare your answer with the correct one above

Question

Find the length of the parametric curve described by

$x = 2\sin t$

$y = 2\cos t$

from to $\pi$ .

Answer

There are several ways to solve this problem, but the most effective would be to notice that we can derive the following-

$x^2 = 4\sin^2t$

$y^2 = 4\cos^2t$

Hence

$x^2+y^2 = 4\sin^2t + 4\cos^2t = 4$

Therefore our curve is a circle of radius $\sqrt4=2$ , and it's circumfrence is $4\pi$ . But we are only interested in half that circumfrence ( is from to $\pi$ , not $2\pi$ .), so our answer is $2\pi$ .

Alternatively, we could've found the length using the formula

$l = \int_{t_0}^{t_1}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$ .

Compare your answer with the correct one above

Question

Find the coordinates of the curve function

when .

Answer

To find the coordinates, we set into the curve function.

We get

and thus

Compare your answer with the correct one above

Question

Find the coordinates of the curve function

when $t=\frac{\pi}{2}$

Answer

To find the coordinates, we evaluate the curve function for $t=\pi/2$

As such,

$c(\frac{\pi}{2})=<cos (\frac{\pi}{2}), sin (\frac{\pi}{2}), cos (\frac{\pi}{2})>=<0,1,0>$

Compare your answer with the correct one above

Question

Find the coordinates of the curve function

when

Answer

To find the coordinates, we evaluate the curve function for

As such,

Compare your answer with the correct one above

Question

Find the equation of the line passing through the two points, given in parametric form:

Answer

To find the equation of the line passing through these two points, we must first find the vector between them:

$\vec{v}=\left \langle 1, 6, 2\right \rangle$

This was done by finding the difference between the x, y, and z components for the vectors. (This can be done in either order, it doesn't matter.)

Now, pick a point to be used in the equation of the line, as the initial point. We write the equation of line as follows:

$\vec{r}=\left \langle 1, 2, 3\right \rangle + t\left \langle 1, 6, 2\right \rangle=\left \langle 1+t, 2+6t, 3+2t\right \rangle$

The choice of initial point is arbitrary.

Compare your answer with the correct one above

Question

Find the coordinate of the parametric curve when

$t=\pi/2$

,

Answer

To find the coordinates of the parametric curve we plug in for

$t=\pi/2$ .

$x=sin,(\pi/2)= 1$

$y=(\pi/2)^2=\pi^2/4$

As such the coordinates are

$(1,\pi^2/4)$

Compare your answer with the correct one above

Question

Write the parametric equations of the line that passes through the points $\left ( 0,1,3\right )$ and $\left ( 2,5,8\right )$ .

Answer

First, you must find the vector that is parallel to the line.

This vector is

$v=\left \langle x_1-x_2,y_1-y_2,z_1-z_2\right \rangle$ .

From the points we were given, this becomes

$\left \langle -2,-4,-5\right \rangle$ .

To form the parametric equations, we need to pick a point that lies on the line we want.

The point $\left ( 0,1,3\right )$ is used.

The vector form of the line is from the following equation

$r=\left ( x_1,y_1,z_1\right )+t\left \langle v\right \rangle=\left ( 0,1,3\right )+t\left \langle -2,-4,-5\right \rangle$ .

We then rewrite each expression in terms of the variables x, y, and z.

Compare your answer with the correct one above

Question

Evaluate the line integral $\int_C f(x,y)ds$ of the function

over the line segment from to

Answer

Evaluate the line integral $\int_C f(x,y)ds$ using the function

$f(x,y)=y+\cos(\pi x)$ over the line segment from to

Define the Parametric Equations to Represent

The points given lie on the line $y = \frac{5}{4}x$ . Define the parameter , then can be written $y =\frac{5}{4}t$ . Therefore, the parametric equations for are:

$y(t)=\frac{5}{4}t$

_________________________________________________________________

The line integral of a function along the curve with the parametric equation and with $a\leq t\leq b$ is defined by:

$\int_C f(x,y)ds = \int_a^bf(x(t),y(t))\left | \vec{r'}(t)\right |dt$ (1)

Where $\vec{r'(t)}$ is the vector derivative of the vector $\vec{r}(t)=\left<x(t),y(t)\right>$ , therefore $\left | \vec{r'(t)}\right |$ is simply the magnitude of the vector derivative.

______________________________________________________________

Write the vector $\vec{r(t)}$ :

$\vec{r(t)}=\left<t, \frac{5}{4}t\right>$

Differentiate,

$\vec{r'(t)}=\left<1,\frac{5}{4}\right>$

The absolute value (magnitude) of this vector is:

$\left | \vec{r'}(t)\right |=\left | \left<1,\frac{5}{4 }\right>\right |=\sqrt{1^2 +\left(\frac{5}{4} \right )^2}=\frac{\sqrt{41}}{4}$

Write the function in terms of the parameter :

$f(x(t),y(t))=f\left(t,\frac{5}{4}t\right) =\frac{5}{4}t+\cos(\pi t)$

Insert everything into Equation (1) noting that the limits of integration will be $0\leq t \leq 4$ due to the fact that the parameter varies from to over the line segment we are integrating over.

$\int_C(y+\cos(\pi x))ds=\int_0^4\left(\frac{5}{4}t+\cos(\pi t)\right)\frac{\sqrt{41}}{4}dt$

$=\frac{\sqrt{41}}{4}\left[\frac{5}{8}t^2+\frac{1}{\pi}\sin(\pi t) \right ]_{t=0}^{t=4}$

$=\frac{\sqrt{41}}{4}\left[\frac{5}{8}(16)+\frac{1}{\pi }\sin(4 \pi) -0-\frac{1}{\pi}\sin(0)\right ]$

$=\frac{5\sqrt{41}}{2}$

Compare your answer with the correct one above

Question

Find the coordinates of the curve function

when $t=\frac{\pi}{2}$

Answer

To find the coordinates, we evaluate the curve function for $t=\pi/2$

As such,

$c(\frac{\pi}{2})=<cos (\frac{\pi}{2}), sin (\frac{\pi}{2}), cos (\frac{\pi}{2})>=<0,1,0>$

Compare your answer with the correct one above

Question

Find the coordinates of the curve function

when

Answer

To find the coordinates, we evaluate the curve function for

As such,

Compare your answer with the correct one above

Question

Find the length of the parametric curve described by

$x = 2\sin t$

$y = 2\cos t$

from to $\pi$ .

Answer

There are several ways to solve this problem, but the most effective would be to notice that we can derive the following-

$x^2 = 4\sin^2t$

$y^2 = 4\cos^2t$

Hence

$x^2+y^2 = 4\sin^2t + 4\cos^2t = 4$

Therefore our curve is a circle of radius $\sqrt4=2$ , and it's circumfrence is $4\pi$ . But we are only interested in half that circumfrence ( is from to $\pi$ , not $2\pi$ .), so our answer is $2\pi$ .

Alternatively, we could've found the length using the formula

$l = \int_{t_0}^{t_1}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$ .

Compare your answer with the correct one above

Question

Find the coordinates of the curve function

when .

Answer

To find the coordinates, we set into the curve function.

We get

and thus

Compare your answer with the correct one above

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