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anonymous

  • 5 years ago

. 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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  1. anonymous
    • 5 years ago
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    y-e^(2t) + 1 and x = e^t

  2. anonymous
    • 5 years ago
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    Take the derivative of your function and plug in t = 2. The answer is your slope.

  3. anonymous
    • 5 years ago
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    so would it be 2e^2t +1 as my derivative for y which is 110.2?

  4. anonymous
    • 5 years ago
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    Derivative of a constant is 0.

  5. anonymous
    • 5 years ago
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    it would be 2et^(2t-1).

  6. anonymous
    • 5 years ago
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    the 2t is pulled out in front and you must subtract 1 from your exponent.

  7. anonymous
    • 5 years ago
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    No -1 in the exponent.

  8. anonymous
    • 5 years ago
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    well its y= e^(2t) + 1

  9. amistre64
    • 5 years ago
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    umm.... I would take a gander and say: 2 e^(2t)

  10. anonymous
    • 5 years ago
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    ok so once i find the derivative for y i just plug in 2 and thats the answer

  11. anonymous
    • 5 years ago
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    yes

  12. anonymous
    • 5 years ago
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    ok i was confused about what i would need to do with the x=e^t part

  13. anonymous
    • 5 years ago
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    derivative of e^t is e^t dt

  14. anonymous
    • 5 years ago
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    in this case you can ignore dt since it will be 1.

  15. anonymous
    • 5 years ago
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    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)?

  16. anonymous
    • 5 years ago
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    i mean y'/x'

  17. anonymous
    • 5 years ago
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    for the final answer?

  18. anonymous
    • 5 years ago
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    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.

  19. anonymous
    • 5 years ago
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    ok

  20. anonymous
    • 5 years ago
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    If you want to check your derivative use wolframalpha

  21. anonymous
    • 5 years ago
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    i mean i know the derivatives are right i just dont understand what i do for the problem

  22. anonymous
    • 5 years ago
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    Ok. Yes all you do is plug in the value given for t.

  23. anonymous
    • 5 years ago
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    ok but for both equations or just the y

  24. anonymous
    • 5 years ago
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    i mean i know the derivatives are right i just dont understand what i do for the problem

  25. anonymous
    • 5 years ago
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    i mean i know the derivatives are right i just dont understand what i do for the problem

  26. anonymous
    • 5 years ago
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    thats the part i dont get

  27. anonymous
    • 5 years ago
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    Ok so your saying your not sure since the equation is in terms of X

  28. amistre64
    • 5 years ago
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    is this a f(x,y) type thing?

  29. anonymous
    • 5 years ago
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    It should work the same way I do believe.

  30. anonymous
    • 5 years ago
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    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 .

  31. anonymous
    • 5 years ago
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    those are the two problems we have were doing stuff with polar coordinates and all that so i'm not sure

  32. anonymous
    • 5 years ago
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    i'm confused on both i just don't get what they want for the answer i guess

  33. anonymous
    • 5 years ago
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    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.

  34. anonymous
    • 5 years ago
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    The derivatives Y-value at x = 1 that is.

  35. anonymous
    • 5 years ago
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    so for the first one since x= e^t would the answer be y=x^2 +1

  36. anonymous
    • 5 years ago
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    No your derivative will be in terms of X, not Y. So it will be x = your derivative

  37. anonymous
    • 5 years ago
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    not y =

  38. anonymous
    • 5 years ago
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    The derivative is simply e^t

  39. anonymous
    • 5 years ago
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    x = e^t

  40. anonymous
    • 5 years ago
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    ya i get that

  41. anonymous
    • 5 years ago
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    but i dont know what they want me to do?

  42. anonymous
    • 5 years ago
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    bc it needs to be in cartesian form not polar

  43. anonymous
    • 5 years ago
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    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.

  44. anonymous
    • 5 years ago
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    Ah ok well you've gone above my head. We're just about to cover that stuff.

  45. anonymous
    • 5 years ago
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    The polar stuff.

  46. anonymous
    • 5 years ago
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    pjschlotter can you help?

  47. anonymous
    • 5 years ago
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    Yes he can he's very smart :) lol

  48. anonymous
    • 5 years ago
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    i guess i'll just repost it to see if someone else can help me thanks tho

  49. anonymous
    • 5 years ago
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    i guess i'll just repost it to see if someone else can help me thanks tho

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