Question

Parametric equation with calculus help.

I am supposed to calculate the speed of an object that has these two parametric equations
x(t)= e^t + e^-t
y(t) = e^-t

Answer 1

\[\Large\rm \frac{dy}{dx}=\frac{(dy/dt)}{(dx/dt)}\]

Answer 2

And then speed is uhhhh, directionless, yes?
So we need the magnitude of whatever velocity we get.

Answer 3

So looks like we need to take a couple derivatives :o

Answer 4

I tried that and I got derivative that ended up looking like this:
dy/dt = (-e^-t)/(e^t -e^-t).

However I'm not getting the right answers even though I did the exact same thing you brought up.

Answer 5

Damnit I mean dy/dx not dy/dt

Answer 6

\[\Large\rm \frac{dy}{dx}=\frac{(dy/dt)}{(dx/dt)}=\frac{-e^{-t}}{e^t-e^{-t}}\]And that's not working out?
Let's um.. try to clean it up a bit maybe. I dunno.

Answer 7

yeah that's exactly what I got

Answer 8

Doesn't work though so idk.

Answer 9

\[\Large\rm \frac{-e^{-t}}{e^t-e^{-t}}\color{royalblue}{\left(\frac{e^t+e^{-t}}{e^t+e^{-t}}\right)}=\frac{e^{-2t}-1}{e^{2t}-e^{-2t}}\]Ehh I guess that doesn't look at that much better lol.

Maybe we just need to take the negative off the front, to show that a SPEED should be directionless, no negative direction.\[\Large\rm \frac{e^{-t}}{e^t-e^{-t}}\]Ahhh I dunno >.< Hmmm

Answer 10

Hmm that's actually not a bad idea because in physics speed is a scalar not a vector, meaning it can't be negative.

Answer 11

The question is asking for speed, but it's for a calculus class so idk maybe they're assuming the student knows that speed is a scalar?

Answer 12

I know that when you get into multivariable calculus you're expected to know that speed is a scalar.

I can't remember before that though.

Answer 13

You're trying to enter this into a site or something?

Answer 14

Yeah exactly.

Answer 15

yeah in physics 1, speed is scalar, and velocity is vector

Answer 16

yeah exactly nincompoop.

Answer 17

in math, they don't give a flip

Answer 18

you define them

Answer 19

It's the only thing I can think of that would make sense, I mean I did the (dy/dt)/(dx/dt) thng and I got the exact same equation zepdrix did.

Answer 20

Have you entered other problems without having any issues?
Like you're comfortable with the input that you need for whatever site you're using?

Sometimes they expect exp{} for the exponential base e.

Answer 21

For the most part yes, but sometimes the website can be dumb so it could just be the website. However there is another problem that I need to raise with the equation.

Answer 22

The problem I am tackling is specifically about calculating minimums and maximums on the interval [0,3] but as you can see with the equation we derived, if you set t = 0 the derivative does not exist at that point because the denominator is 0. I even calculated the critical numbers for the derivative and got 0 as the only critical number. So that's another problem.

Answer 23

what website?

Answer 24

erm, idk I found this. not sure if it's the right equation?

Answer 25

webassign.

Answer 26

I did some google stuff... are we supposed to use this instead for parametric?\[\Large\rm v(t)=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\]

Answer 27

i knew it laughing out loud

Answer 28

except replace v(t) with dp(t) :))

Answer 29

Idk possibly.

Answer 30

We never learned that though so that would be odd if that were the case.

Answer 31

arc length uses pythagorean

Answer 32

?

Answer 33

https://www.boundless.com/calculus/differential-equations-parametric-equations-and-sequences-and-series/parametric-equations-and-polar-coordinates/arc-length-and-speed/

Answer 34

I'll try this new equation and see if it works.

Answer 35

look at the three key points, and you will be set for life

Answer 36

Lol that equation worked, the Pythagorean one.

Answer 37

Thanks guys, strange though because we were never taught this but I'm going to save the pages you gave me and study them.

Answer 38

Idk who I shall give my medal too.....

Answer 39

I'll flip a coin heads goes to zepdrix, tails goes to nincompoop

Answer 40

Zepdrix wins

Answer 41

I don't want medals
a masters degree is enough for me to aim for

Answer 42

Ha, thanks for the help I'm going to close this now.