. Calculate the slope of the tangent line to the curve at the point where t = 2:
y = e 2t + 1 and x = e t .

- anonymous

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- anonymous

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

- anonymous

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

- anonymous

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

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## More answers

- anonymous

Derivative of a constant is 0.

- anonymous

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

- anonymous

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

- anonymous

No -1 in the exponent.

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

derivative of e^t is e^t dt

- anonymous

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

- anonymous

ok so i just plug in 2 for that and it would be e^2 so do i i have e^2 divided by my answer for y'(2)?

- anonymous

i mean y'/x'

- anonymous

for the final answer?

- anonymous

your answer would just be e^2 I think. What that is saying is the derivative at the point t=2 would be e^2, since the derivative is e^t. not 100% sure on that tho now i'm confusing myself I think.

- anonymous

ok

- anonymous

If you want to check your derivative use wolframalpha

- anonymous

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

- anonymous

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

- anonymous

ok but for both equations or just the y

- anonymous

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

- anonymous

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

- anonymous

thats the part i dont get

- anonymous

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

- amistre64

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

- anonymous

It should work the same way I do believe.

- anonymous

1. Eliminate the parameter to find the Cartesian equation for the curve:
y = (e ^2t) + 1 and x = e^ t .
2. Calculate the slope of the tangent line to the curve at the point where t = 2:
y = (e ^2t) + 1 and x = e ^t .

- anonymous

those are the two problems we have were doing stuff with polar coordinates and all that so i'm not sure

- anonymous

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

- anonymous

Well the first one is simple. The derivative of a function is a function that is used to calculate the slope of it's antiderivative at that point. So lets say the slope of a function at point x = 1 is 2. The derivative's Y value at 1 would be 2, since the derivative allows you to calculate the slope at a given point.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

not y =

- anonymous

The derivative is simply e^t

- anonymous

x = e^t

- anonymous

ya i get that

- anonymous

but i dont know what they want me to do?

- anonymous

bc it needs to be in cartesian form not polar

- anonymous

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.

- anonymous

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

- anonymous

The polar stuff.

- anonymous

pjschlotter can you help?

- anonymous

Yes he can he's very smart :) lol

- anonymous

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

- anonymous

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

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