Parametric Curves - Calculus 3

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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:

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

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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.

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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.

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

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

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

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

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