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y-e^(2t) + 1 and x = e^t

Take the derivative of your function and plug in t = 2. The answer is your slope.

so would it be 2e^2t +1 as my derivative for y which is 110.2?

Derivative of a constant is 0.

it would be 2et^(2t-1).

the 2t is pulled out in front and you must subtract 1 from your exponent.

No -1 in the exponent.

well its y= e^(2t) + 1

umm.... I would take a gander and say:
2 e^(2t)

ok so once i find the derivative for y i just plug in 2 and thats the answer

yes

ok i was confused about what i would need to do with the x=e^t part

derivative of e^t is e^t dt

in this case you can ignore dt since it will be 1.

i mean y'/x'

for the final answer?

ok

If you want to check your derivative use wolframalpha

i mean i know the derivatives are right i just dont understand what i do for the problem

Ok. Yes all you do is plug in the value given for t.

ok but for both equations or just the y

i mean i know the derivatives are right i just dont understand what i do for the problem

i mean i know the derivatives are right i just dont understand what i do for the problem

thats the part i dont get

Ok so your saying your not sure since the equation is in terms of X

is this a f(x,y) type thing?

It should work the same way I do believe.

i'm confused on both i just don't get what they want for the answer i guess

The derivatives Y-value at x = 1 that is.

so for the first one since x= e^t would the answer be y=x^2 +1

No your derivative will be in terms of X, not Y. So it will be x = your derivative

not y =

The derivative is simply e^t

x = e^t

ya i get that

but i dont know what they want me to do?

bc it needs to be in cartesian form not polar

Ok. you take that derivative (x = e^t) and plug in t=2. You get x = e^2. Your slope at t = 2 is e^2.

Ah ok well you've gone above my head. We're just about to cover that stuff.

The polar stuff.

pjschlotter can you help?

Yes he can he's very smart :) lol

i guess i'll just repost it to see if someone else can help me thanks tho

i guess i'll just repost it to see if someone else can help me thanks tho