Products
Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

## Q. 1.4

Describe and sketch the curve given by the vector equation $\mathbf{f}(r)=\left(1-r^{4}\right) \mathbf{i}+r^{2} \mathbf{j}$.

## Verified Solution

First, we note that the parameter in this problem is r instead of t .This makes absolutely no difference since, as in the case of the variable of integration, the parameter is a “dummy” variable. Now to get a feeling for the shape of this curve, we display in Table 1, values of x and y for various values of r .Plotting some of these points leads to the sketch in Figure 3. To write the Cartesian equation for this curve we square both sides of the equation $y=r^{2}$ to obtain $y^{2}=r^{4}$ and $x=1-r^{4}=1-y^{2}$ which is the equation of the parabola sketched in Figure 4. Note that this curve is not the same as the curve sketched in Figure 3 since the parametric representation $y=r^{2}$ requires that y be nonnegative. Thus the curve described by (6)

$\mathbf{f}(r)=\left(1-r^{4}\right) \mathbf{i}+r^{2} \mathbf{j}$

is only the upper half of the parabola described by the equation $x=1-y^{2}$.

 r 0 ±$\frac{1}{2}$ ±1 ±$\frac{3}{2}$ ±2 x = 1 – r⁴ 1 $\frac{15}{16}$ 0 –$\frac{65}{16}$ -15 y = r² 0 $\frac{1}{4}$ 1 $\frac{9}{4}$ 4