Question

. Calculate the slope of the tangent line to the curve at the point where t = 2:

y = e 2t + 1 and x = e t .

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Derivative of a constant is 0.

Answer 5

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

Answer 6

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

Answer 7

No -1 in the exponent.

Answer 8

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

Answer 9

umm.... I would take a gander and say:

2 e^(2t)

Answer 10

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

Answer 11

yes

Answer 12

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

Answer 13

derivative of e^t is e^t dt

Answer 14

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

Answer 15

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)?

Answer 16

i mean y'/x'

Answer 17

for the final answer?

Answer 18

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.

Answer 19

ok

Answer 20

If you want to check your derivative use wolframalpha

Answer 21

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

Answer 22

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

Answer 23

ok but for both equations or just the y

Answer 24

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

Answer 25

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

Answer 26

thats the part i dont get

Answer 27

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

Answer 28

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

Answer 29

It should work the same way I do believe.

Answer 30

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 .

Answer 31

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

Answer 32

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

Answer 33

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.

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

not y =

Answer 38

The derivative is simply e^t

Answer 39

x = e^t

Answer 40

ya i get that

Answer 41

but i dont know what they want me to do?

Answer 42

bc it needs to be in cartesian form not polar

Answer 43

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.

Answer 44

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

Answer 45

The polar stuff.

Answer 46

pjschlotter can you help?

Answer 47

Yes he can he's very smart :) lol

Answer 48

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

Answer 49

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