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Math Help - Am I Right? (Differentiation)

  1. #1
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    Am I Right? (Differentiation)

    1. Find the equation of the tangent at the given points for: f(x) = x2 + 3x, (2, 10)
    I did:
    f'(x) = 2x + 3
    Let x = 2
    --> m = 7
    y - 10 = 7(x - 2)
    y = 7x - 4

    2. Find the equation of the tangent to the curve y = x3 - 9x2 + 20x - 8 at the point (1, 4). At what points of the curve is the tangent parallel to the line y = -4x + 3?
    First, I used the method from the previous question to obtain the equation for the parallel line. For the next part of the question, I equated the gradient function to the gradient of the parallel line and solved for x, substituting the newly found x values into the original curve equation to get my y-coordinates.

    3. A model plane flying level at 250 m above the ground suddenly dives... blahablahhal... h(t) = 8t2 - 80t + 250. Find the rate at which the plane is losing height at t = 3.
    I simply substituted t = 3 into the original equation and solved, giving my answer as 82 ms-1. Is this question as simple as that?

    4. A car starts from rest and moves a distance s (m) in t (s), where s = (1/6)t3 + (1/4)t2. What is the acceleration when t = 2?
    I know that velocity = ds/dt and acceleration = dv/dt, so I found that s' = (1/2)t2 + (1/2)t and v' = t + (1/2). Then, to answer the question, I simply substituted t = 2 into the equation for acceleration and solved, giving my (5/2) ms-2.


    Thanks in advance. Your feedback is greatly appreciated.
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  2. #2
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    Re: Am I Right? (Differentiation)

    Quote Originally Posted by Fratricide View Post
    1. Find the equation of the tangent at the given points for: f(x) = x2 + 3x, (2, 10)
    I did:
    f'(x) = 2x + 3
    Let x = 2
    --> m = 7
    y - 10 = 7(x - 2)
    y = 7x - 4
    Yes,, that is correct.

    2. Find the equation of the tangent to the curve y = x3 - 9x2 + 20x - 8 at the point (1, 4). At what points of the curve is the tangent parallel to the line y = -4x + 3?
    First, I used the method from the previous question to obtain the equation for the parallel line. For the next part of the question, I equated the gradient function to the gradient of the parallel line and solved for x, substituting the newly found x values into the original curve equation to get my y-coordinates.
    Sounds good. Why have you not done it?

    3. A model plane flying level at 250 m above the ground suddenly dives... blahablahhal... h(t) = 8t2 - 80t + 250. Find the rate at which the plane is losing height at t = 3.
    I simply substituted t = 3 into the original equation and solved, giving my answer as 82 ms-1. Is this question as simple as that?
    No, it's not. Putting h= 3 into h(t) gives h(3)= 82 meters- not "meters per second", because, as you were told, h(t) is the height of the plane at t seconds, NOT the rate at which that height is changing.

    4. A car starts from rest and moves a distance s (m) in t (s), where s = (1/6)t3 + (1/4)t2. What is the acceleration when t = 2?
    I know that velocity = ds/dt and acceleration = dv/dt, so I found that s' = (1/2)t2 + (1/2)t and v' = t + (1/2). Then, to answer the question, I simply substituted t = 2 into the equation for acceleration and solved, giving my (5/2) ms-2.
    Yes, that is correct. Why didn't you do that for the previous question?

    Thanks in advance. Your feedback is greatly appreciated.
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  3. #3
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    Re: Am I Right? (Differentiation)

    Quote Originally Posted by HallsofIvy View Post
    Sounds good. Why have you not done it?
    I have, I just didn't include it for brevity's sake.
    Here it is:
    dy/dx = 3x2 - 18x + 20
    Let x = 1
    --> m = 5
    y - 4 = 5(x - 1)
    y = 5x - 1

    y = -4x + 3
    m = -4
    Let 3x2 - 18x + 20 = -4
    3x2 - 18x + 24 = 0
    (3x - 12)(x - 2) = 0
    x = 2, 4

    Let x = 2
    y = 2(2)3 - 9(2)2 + 20(2) - 8
    y = 4

    Let x = 4
    y = 2(4)3 - 9(4)2 + 20(4) - 8
    y = -8

    --> Tangent is parallel to the line y = -4x + 3 at points (2,4) and (4,-8).

    Quote Originally Posted by HallsofIvy View Post
    Why didn't you do that for the previous question?
    Ahhhh, I see. So I need to use the same method as in 4, substituting t = 3 into v'?
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    Re: Am I Right? (Differentiation)

    Ignore that last part of my previous post. I'm pretty sure that for question three I need to find the velocity (not acceleration) at the given time.
    As such, I did:
    h'(t) = 16t - 80
    Let t = 3
    h'(t) = -32 ms-1

    Am I on the right track? (With this question and the workings out in my previous post.)
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  5. #5
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    Re: Am I Right? (Differentiation)

    I'm pretty sure what I've done is correct, but can anyone confirm it for me?
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  6. #6
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    Re: Am I Right? (Differentiation)

    If you get h'(3) = -32, then the plane is descending at 32m/s.
    Thanks from Fratricide
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