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

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

Mathematics
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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?

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Other answers:

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.
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)?
i mean y'/x'
for the final answer?
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.
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.
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 .
those are the two problems we have were doing stuff with polar coordinates and all that so i'm not sure
i'm confused on both i just don't get what they want for the answer i guess
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.
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

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