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Math Help - odd questions i'd like some advice on

  1. #1
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    odd questions i'd like some advice on

    just had a quiz today and teacher threw on some questions that stumped me quite a bit and i'm trying to understand what i was supposed to do

    1:
    find the tangent line to the curve y = x-8/x+8 that is parallel to x-y = 8

    so i differentiate the y='s equation to get (x+8) - (x-8) / (x+8)^2

    that's where i'm stuck at and i don't get the parallel part

    2: position of a particle is found by s = t^3 - 1.5t^2 - 2t. how long does it take to reach a velocity of 166 m/s

    what i did was take the derivative of each individual term by the sum rule to get 3t^2 - 3t - 2

    how do i find out when velocity reaches 166 with that equation?
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  2. #2
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    Quote Originally Posted by deltemis View Post
    1:
    find the tangent line to the curve y = x-8/x+8 that is parallel to x-y = 8
    so i differentiate the y='s equation to get (x+8) - (x-8) / (x+8)^2
    The derivative is y'=\frac{16}{(x+8)^2}.
    The slope you are looking for is 1 and that happens if x=-4.
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  3. #3
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    hmm i differentiated wrong. i'm wondering how you found the slope and at what x it occurs?

    also does anybody know how to get my second problem? i'm having no luck with whatever i try
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  4. #4
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    Quote Originally Posted by deltemis View Post
    [snip]

    2: position of a particle is found by s = t^3 - 1.5t^2 - 2t. how long does it take to reach a velocity of 166 m/s

    what i did was take the derivative of each individual term by the sum rule to get 3t^2 - 3t - 2

    how do i find out when velocity reaches 166 with that equation?
    ds/dt = v = 3t^2 - 3t - 2. Substitute v = 166 and solve for t.

    Quote Originally Posted by deltemis View Post
    hmm i differentiated wrong. i'm wondering how you found the slope and at what x it occurs?[snip]
    x-y = 8 is a line and has a gradient of 1. Tangent parallel to x-y = 8 means gradient of tangent = 1. Substitute y' = 1 and solve for x.
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