Converting Parametric Equations

[LongTermStudent]

I have an upcoming exam, and I'm having trouble grasping some concepts. The things that are currently perplexing me are parametric equations and rectangular equations and converting between the two. I have a problem like this

Given the parametric equations x = e^(-t) + 1 and y = e^(-2t) - 3, find the corresponding rectangular equation and sketch the curve of orientation.

To eliminate the parameter, I set the x equation equal to t as such:
t = -ln(x-1)

and then substituted this value into the y equation to get:
y = e^[ln(x - 1)] - 3
and got y = x - 4 as my final equation, but I'm not so confident in this answer.

Then I'm told to find a set of parametric equatiions for the rectangular equation (x - 2)² + y² = 9

I set x equal to t and soved for y and ended up with
y = sqrt(t²-4t+13)

This seems like only half the answer to me, but I barely know where to start, let alone how to continue. I'm sure I'll be kicking myself when I finally figure these concepts out, but I can't for the life of me wrap my head around any of it, and I have no access to a tutor between now and my exam. Any help/guidance is appreciated.

Edit: I see now I may have posted this in the wrong forum. My bad.

 

[bob1182006]

for the first one, your substitution was wrong, you should get e^(-2t) = e^(-2(-ln(x-1))) which doesn't go down to just (x-1)

for the second one I think you get down to y^2=t^2+4t+13 right? then you square root both sides? you should get y=+ or - sqrt(...) since both +sqrt(...) and -sqrt(...) will satisfy that equation.

 

[tacman]

Hi there,

First of all, don't worry - it's completely normal to struggle with certain concepts before an exam. It's important to keep practicing and seeking help when needed, so good for you for reaching out for guidance.

Let's start with converting parametric equations to rectangular equations. The process you followed for the first problem is correct. To eliminate the parameter t, you need to solve for t in terms of x and then substitute that value into the other equation. So, t = -ln(x-1) is correct.

Next, when you substituted that value into the y equation, you made a small mistake. Remember, when raising e to a power, the inverse operation is taking the natural logarithm (ln). So, instead of e^[ln(x-1)], it should be just (x-1). So, your final equation should be y = x - 4. This is the correct rectangular equation for the given parametric equations.

Now, for the second problem, you're on the right track. You need to set x equal to t and solve for y. But, remember that when you square both sides of an equation, you need to include a positive and negative solution. So, when you take the square root, you should have two possible equations: y = sqrt(t²-4t+13) and y = -sqrt(t²-4t+13). These are the two sets of parametric equations for the given rectangular equation.

Finally, to sketch the curve of orientation, you can plot points by choosing different values for t and then plugging them into the parametric equations. For example, if t = 0, then x = 2 and y = 3. So, one point on the curve would be (2, 3). Similarly, you can choose other values for t and plot more points to get a better understanding of the curve.

I hope this helps and good luck on your exam! Remember to keep practicing and don't hesitate to seek help if needed.