# (a) Eliminate the parameter t and express y explicitly in terms of x for the equations x = \frac...

## Question:

(a) Eliminate the parameter {eq}t {/eq} and express {eq}y {/eq} explicitly in terms of {eq}x {/eq} for the equations {eq}x = \frac {1 - t^2}{1 + t^2} {/eq} and {eq}y = \frac {2t}{1 + t^2} {/eq} .

(b) Find the projection of {eq}\vec b {/eq} along {eq}\vec a {/eq} for the vectors {eq}\vec a = (0 , 3 , 4 ) {/eq} and {eq}\vec b = ( -7 , 10 , 9 ) {/eq} .

(c) Eliminate the parameter {eq}t {/eq} and express {eq}y {/eq} as a function of {eq}x {/eq} for the equations {eq}x = e^t {/eq} and {eq}y = e^{-t} {/eq} .

(d) Given the vectors {eq}\vec a = ( 3 , -11 , 2 ) {/eq} and {eq}\vec b = ( 3 , -4 , 9 ) {/eq} , find the unit vector {eq}\vec u {/eq} in the direction of the vector {eq}\vec a - \vec b {/eq} .

## Curves and Vectors:

In this odd back and forth we will bounce between working with parametric equations and vectors. We will deal with the parametric equations as they come up. For the vectors, recall that the projection of the vector {eq}\vec v {/eq} along the vector {eq}\vec u {/eq} is determined by

{eq}\begin{align*} \text{proj}_{\vec u} \vec v ****= \frac{\vec v \cdot \vec u}{| \vec u | ^2}\ \vec u \end{align*} {/eq}

(a)

The tricky one is right up front here. Plotting the function reveals it to be the unit circle. So we can eliminate the parameter {eq}t {/eq} by taking the sum of the squares:

{eq}\begin{align*} x^2 + y^2 &= \left( \frac {1-t^2}{1 + t^2}\right)^2 + \left( \frac {2t}{1 + t^2} \right)^2 \\ &= \frac{1 - 2t^2 + t^4 + 4t^2 }{(1+t^2)^2} \\ &= \frac{1 + 2t^2 + t^4 }{(1+t^2)^2} \\ &= \frac{(1+t^2)^2}{(1+t^2)^2} \\ &= 1 \end{align*} {/eq}

So the curve is

{eq}\begin{align*} x^2 + y^2 &= 1 \end{align*} {/eq}

as expected.

(b)

We have the vectors {eq}\vec a = \left< 0 , 3 , 4 \right> {/eq} and {eq}\vec b = \left< -7 , 10 , 9 \right> {/eq}. We have

{eq}\begin{align*} \vec a \cdot \vec b &= \left< 0 , 3 , 4 \right> \cdot \left< -7 , 10 , 9 \right> \\ &= 0 + 30 + 36 \\\ &= 66 \end{align*} {/eq}

and

{eq}\begin{align*} | \vec a | &= \sqrt{\left( 0 \right)^2+\left( 3 \right)^2+\left(4 \right)^2} \\ &= 5 \end{align*} {/eq}

And so the projection of the vector {eq}\vec b {/eq} along the vector {eq}\vec a {/eq} is

{eq}\begin{align*} \text{proj}_{\vec a} \vec b &= \frac{\vec b \cdot \vec a}{| \vec a | ^2}\ \vec a \\ &= \frac{66}{5^2} \ \left< 0 , 3 , 4 \right> \\ &= \left< 0,\ \frac{198}{25},\ \frac{264}{25} \right> \end{align*} {/eq}

(c)

This one is nice and easy. We have

{eq}x = e^t {/eq} and {eq}y = e^{-t} {/eq}

so

{eq}\begin{align*} y &= \frac1{e^t} \\ &= \frac1x \end{align*} {/eq}

(d) Lastly, we have the vectors {eq}\vec a = \left<3 , -11 , 2 \right> {/eq} and {eq}\vec b =\left< 3 , -4 , 9 \right> {/eq}. Then

{eq}\begin{align*} \vec u &= \vec a - \vec b \\ &= \left< 3 , -11 , 2 \right> - \left< 3 , -4 , 9 \right> \\ &= \left< 0, -7, -7 \right> \\ &= 7 \left< 0,-1,-1 \right> \end{align*} {/eq}

So the unit vector is

{eq}\begin{align*} \hat u &= \frac{7 \left< 0,-1,-1 \right>}{|7 \left< 0,-1,-1 \right>|} \\ &= \frac{\left< 0,-1,-1 \right>}{\sqrt2} \\ &= \left< 0,\ - \frac{\sqrt2}{2},\ - \frac{\sqrt2}{2} \right> \end{align*} {/eq}