Chapter 11.1, Problem 18E

To determine

To sketch the curve over given t-interval, indicating the direction in which it is traced.

Explanation of Solution

Given:

The given parametric equations are

  $\mathrm{x}={\mathrm{t}}^2+2\mathrm{t},\mathrm{y}={\mathrm{t}}^2-2\mathrm{t}+3-2\le \mathrm{t}\le 3$

Calculation:

For different values of t we find and locate different points. By joining these points we get our curve.

The sketch of the parametric equations is as follows:

  

To determine

To identify the leftmost point.

Answer to Problem 18E

  $(-1,-6)$

Explanation of Solution

Given:

The given parametric equations are

  $\mathrm{x}={\mathrm{t}}^2+2\mathrm{t},\mathrm{y}={\mathrm{t}}^2-2\mathrm{t}+3-2\le \mathrm{t}\le 3$

Calculation:

The leftmost point is $(-1,-6)$

To determine

To sketch the curve over given t-interval, indicating the direction in which it is traced.

Answer to Problem 18E

  $(-1,6)$

Explanation of Solution

Given:

The given parametric equations are

  $\mathrm{x}={\mathrm{t}}^2+2\mathrm{t},\mathrm{y}={\mathrm{t}}^2-2\mathrm{t}+3-2\le \mathrm{t}\le 3$

Calculation:

To find the leftmost point we can use

  $\begin{array}{l}\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}=0\\ 2\mathrm{t}+2=0\\ \mathrm{t}=-1\end{array}$

At $\mathrm{t}=-1$ the value of x is minimum.

And that is $\mathrm{x}={(-1)}^2+2(-1)=-1$

And also,

  $\mathrm{y}={(-1)}^2-2(-1)+3=6$

So, the left most point by using derivative is $(-1,6)$