Solutions to Problem Set 30

Math 211-03

11-20-2017

[Parametric equations]

1. Find parametric equations for the curve $y = x^2$ for $-1 \le x \le
   1$ .

Set the independent variable equal to the parameter, so $x = t$ . Plug this into the x-y equation to obtain $y = t^2$ . The parametric equations are

$$x = t, \quad y = t^2, \quad -1 \le t \le 1.\quad\halmos$$


2. Find parametric equations for the curve $x = y^2 - 3 y - 4$ for $-2
   \le y \le 5$ .

Set the independent variable equal to the parameter, so $y = t$ . Plug this into the x-y equation to obtain $x = t^2 - 3 t - 4$ . The parametric equations are

$$x = t^2 - 3 t - 4, \quad y = t, \quad -2 \le t \le 5.\quad\halmos$$


3. Find the x-y equation for the curve given by the parametric equations

$$x = t + 1, \quad y = t^2 + 2 t + 4.$$

Note that

$$y = t^2 + 2 t + 4 = (t^2 + 2 t + 1) + 3 = (t + 1)^2 + 3 = x^2 + 3.$$

Alternatively, you can solve the first equation for t to get $t = x - 1$ . Plugging this into the y-equation and simplifying gives the same result.


4. Find parametric equations for the segment from the point $P(2, -4)$ to the point $Q(8, -7)$ .

$$(x, y) = (1 - t) (2, -4) + t (8, -7) = [2(1 - t) + 8 t, -4(1 - t) - 7 t] = (2 + 6 t, -4 - 3 t).$$

The equations are

$$x = 2 + 6 t, \quad y = -4 - 3 t, \quad 0 \le t \le 1.\quad\halmos$$


5. Find parametric equations for the vertical line $x = 9$ . (Note that this line is not the graph of a function $y = f(x)$ .)

$$x = 9, \quad y = t, \quad -\infty < t < \infty.\quad\halmos$$


6. Find parametric equations for the circle

$$x^2 + y^2 = 64.$$

Choose your parametrization so that the circle is traced out once in the counterclockwise direction as t goes from 0 to $2 \pi$ .

$$x = 8 \cos t, \quad y = 8 \sin t.\quad\halmos$$


7. Find parametric equations for the circle

$$(x + 5)^2 + (y - 2)^2 = 36.$$

Choose your parametrization so that the circle is traced out once in the counterclockwise direction as t goes from 0 to $2 \pi$ .

Divide the circle equation by 36 and write it as

$$\left(\dfrac{x + 5}{6}\right)^2 + \left(\dfrac{y - 2}{6}\right)^2 = 1.$$

This has the same form as $(\cos
   t)^2 + (\sin t)^2 = 1$ . So I can get a parametrization by

$$\dfrac{x + 5}{6} = \cos t, \quad\hbox{so}\quad x = -5 + 6 \cos t,$$

$$\dfrac{y - 2}{6} = \sin t, \quad\hbox{so}\quad y = 2 + 6 \sin t.$$

The equations are

$$x = -5 + 6 \cos t, \quad y = 2 + 6 \sin t, \quad 0 \le t \le 2 \pi.\quad\halmos$$

Note: If you switch the sine and cosine, the circle is traced out clockwise.


8. Find the x-y equation for the curve given by the parametric equations

$$x = 10 + 3 \cos t, \quad y = -1 + 3 \sin t.$$

Solve the parametric equations for $\cos t$ and $\sin t$ :

$$\cos t = \dfrac{x - 10}{3}, \quad \sin t = \dfrac{y + 1}{3}.$$

Then

$$\left(\dfrac{x - 10}{3}\right)^2 + \left(\dfrac{y + 1}{3}\right)^2 = (\cos t)^2 + (\sin t)^2 = 1.$$

I can simplify the x-y-equation to

$$(x - 10)^2 + (y + 1)^2 = 9.\quad\halmos$$


9. Find the x-y equation for the curve given by the parametric equations

$$x = \sec t, \quad y = \tan t.$$

$$x^2 - y^2 = (\sec t)^2 - (\tan t)^2 = 1.\quad\halmos$$


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