Parametric equations

[mynamesmurph]

Howdy yalls,

Here are the instructions for this problem

The parametric equations are given, graph the plane curve.


x=e^t
y=e^2t - 4

Ordinarily I pick an equation and solve for t
I choose the first equation.

x=e^t
log_ex=log_ee^t
ln(x)=t

Now, I substitute into the second equation.
y=e^2ln(x) - 4

Am I correct thus far?
Assuming I am, I look at ln(x)=t and I see that my values for x must be greater than 1.
Now I start choosing values for t to get x and y values, but my answers are incorrect.

Any help? Thanks a bunch

 

[mynamesmurph]

You know it just occured to me that this was probably a poor way to solve this when I could have just entered in values for t and graphed the x and y values.

That said, I'm still curious about my previous approach and why it failed...

 

[Ishuda]

You know it just occured to me that this was probably a poor way to solve this when I could have just entered in values for t and graphed the x and y values.

That said, I'm still curious about my previous approach and why it failed...

It didn't really. The problem is you stopped to soon, continuing from
y=e^2ln(x) - 4
we note that
$\displaystyle e^{2\space ln(x)}= e^{ln(x^2)} = x^2$
so that
y = x² - 4


You could also have noted that
e^2t = (e^t)² = x²

EDIT: You have run into a problem here which would be covered in a higher level of math and that is imaginary numbers. It turn out that if x is less than zero, the log is an imaginary number [more accurately a complex number which is the sum of a real and imaginary number]. The exponential of a complex number can be a real number and, since the requirement is that y be real, the intermediate step where the exponent is complex, that is x is negative, is allowed.

 

[mynamesmurph]

It didn't really. The problem is you stopped to soon, continuing from
y=e^2ln(x) - 4
we note that
$\displaystyle e^{2\space ln(x)}= e^{ln(x^2)} = x^2$
so that
y = x² - 4

OK, I understand this, I just didn't see it at the time.

You could also have noted that
e^2t = (e^t)² = x²

I don't understand this though. Would you mind explaining?

EDIT: You have run into a problem here which would be covered in a higher level of math and that is imaginary numbers. It turn out that if x is less than zero, the log is an imaginary number [more accurately a complex number which is the sum of a real and imaginary number]. The exponential of a complex number can be a real number and, since the requirement is that y be real, the intermediate step where the exponent is complex, that is x is negative, is allowed.

I'm not sure what you're getting at here. I haven't dealt with logs of negative numbers, but if the number is imaginary I could just graph it on a complex plane, right?

Thanks!

 

[Ishuda]

You could also have noted that
e^2t = (e^t)² = x²

I don't understand this though. Would you mind explaining?

this is just the application of the general rule
u^{a b} = u^{b a} = (u^a)^b = (u^b)^a
In this case u = e, a = 2, and b = t. Also, from the first equation x = e^t

EDIT: You have run into a problem here which would be covered in a higher level of math and that is imaginary numbers. It turn out that if x is less than zero, the log is an imaginary number [more accurately a complex number which is the sum of a real and imaginary number]. The exponential of a complex number can be a real number and, since the requirement is that y be real, the intermediate step where the exponent is complex, that is x is negative, is allowed.

I'm not sure what you're getting at here. I haven't dealt with logs of negative numbers, but if the number is imaginary I could just graph it on a complex plane, right?

To specifically answer your question, yes you could graph it on a complex plane. However, in this case you wouldn't need to do so. For example, let x = -2, then ln(x) = ln(2) + i $\displaystyle {\pi}$, a complex number. But, since
e^{i 2 $\displaystyle {\pi}$} = 1,
e^2x = e^{2 ln(2) + i 2 $\displaystyle {\pi}$} = e^{2 ln(2)} 1 = $\displaystyle e^{ln(2^2)}$ = 2² = (-2)² = x² a real number

Note that that 2 could have been any real positive number so x could be any real negative number.