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Chapter 2

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 – r 1 \frac{15}{16} 0 \frac{65}{16} -15
y = r² 0 \frac{1}{4} 1 \frac{9}{4} 4