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Math Help - [SOLVED] Problem with differenciation

  1. #1
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    Exclamation [SOLVED] Problem with differenciation

    Hi! Please help me with this problem!

    A stone is thrown upwards. It is h metres high after t seconds.
    h=20t-5t2

    So far I found:
    a)When is the stone 15 metres high?

    15= 20t-5t2
    0=4t-t2-3
    (t-3)(t-1)

    So t=1 and 3

    b)When is it highest?

    c)Distance travelled(total)?
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  2. #2
    Super Member fardeen_gen's Avatar
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    Using first derivative test, since we need maximum height(say 'h') in time 't',
    dh/dt = 0
    d(20t - 5t^2)/dt = 0
    20 - 5*2t = 0 [Using constant rule ; dt/dt = 1 ; power rule]
    10t = 20
    t = 2
    Therefore, for t = 2 seconds, height is maximum

    Putting t = 2 in equation for h,

    h = (20*2 - 5 *(2)^2) m
    = (40 - 5*4) m
    = 20 m

    For total distance, since stone reaches a maximum distance of 20 m above ground, total distance is (20 + 20) m, i.e. 40 m.(since stone reaches highest point, then falls back to the ground.)
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  3. #3
    Kai
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    Quote Originally Posted by thegirlwhoneedshelp View Post
    Hi! Please help me with this problem!

    A stone is thrown upwards. It is h metres high after t seconds.
    h=20t-5t2

    So far I found:
    a)When is the stone 15 metres high?

    15= 20t-5t2
    0=4t-t2-3
    (t-3)(t-1)

    So t=1 and 3

    b)When is it highest?

    c)Distance travelled(total)?
    Hi,

    Highest distance is when dh/dt =0

    So u will get 20-10t=0 ...solve for t.

    Second, find h for that t, means it took t seconds to go upward, then covering the same distance back to its initial position

    Hope it helps
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  4. #4
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    Hi! I don't understand how you got the answer in blue below and what the constant rule and the power rule are or have to do with it. Could you explain it for me please?

    Quote Originally Posted by fardeen_gen View Post
    Using first derivative test, since we need maximum height(say 'h') in time 't',
    dh/dt = 0
    d(20t - 5t^2)/dt = 0
    20 - 5*2t = 0 [Using constant rule ; dt/dt = 1 ; power rule]
    10t = 20
    t = 2
    Therefore, for t = 2 seconds, height is maximum

    Putting t = 2 in equation for h,

    h = (20*2 - 5 *(2)^2) m
    = (40 - 5*4) m
    = 20 m

    For total distance, since stone reaches a maximum distance of 20 m above ground, total distance is (20 + 20) m, i.e. 40 m.(since stone reaches highest point, then falls back to the ground.)
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  5. #5
    Moo
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    Hello,
    Quote Originally Posted by thegirlwhoneedshelp View Post
    Hi! I don't understand how you got the answer in blue below and what the constant rule and the power rule are or have to do with it. Could you explain it for me please?
    Constant rule : if a is a constant (here, 5), then you have (a \cdot f(x))'=a \cdot f'(x)

    Power rule : (x^n)'=n \cdot x^{n-1} (here, n=2)
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  6. #6
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    Hi! when I wrote 5t2, I meant 5t squared. Could you tell me what to do then?

    Quote Originally Posted by Kai View Post
    Hi,

    Highest distance is when dh/dt =0

    So u will get 20-10t=0 ...solve for t.

    Second, find h for that t, means it took t seconds to go upward, then covering the same distance back to its initial position

    Hope it helps
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  7. #7
    Super Member fardeen_gen's Avatar
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    Hi! when I wrote 5t2, I meant 5t squared. Could you tell me what to do then?


    Originally Posted by thegirlwhoneedshelp
    Hi! Please help me with this problem!

    A stone is thrown upwards. It is h metres high after t seconds.
    h=20t-5t2

    So far I found:
    a)When is the stone 15 metres high?

    15= 20t-5t2
    0=4t-t2-3
    (t-3)(t-1)

    So t=1 and 3


    If you meant 5t squared, then how did you get the above equation?(Correct me if I am wrong, but by 5t squared, you do mean (5t) * (5t), don't you?)

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