At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

Problem 1I-3 (a) says
\[ \vec{r} = < 2\cos^2 t , \ \sin^2 t> \]
in other words, r is a position vector with x component = 2 cos^2 t and
a y component = sin^2 t
The attached graph show the locus of points we create as we increment t
notice if we start with
\[ x= 2 \cos^2 t \\ y= \sin^2 t \]
and (because we want to use sin^2( t) + cos^2(t) = 1 to simplify the equation) we multiply the 2nd equation by 2 and add the two equations:
\[ x + 2y = 2\cos^2 t + 2 \sin^2 t \\ x+2y= 2(\cos^2 t + \sin^2 t) \\ x+2y= 2 \]
that is the equation of a line. In slope-intercept form
\[ y = - \frac{1}{2} x +1 \]
so we should expect the locus of points to lie on this line (and it does)
because sin and cos are cyclical, we will get cyclical behavior... the point moves back and forth between (0,1) and (2,0)