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    Can someone help me with this question?

    v = 20000te^(-t/3)
    Use this model and calculus to find the value of t when dv/dt = 0

    I can't work out from the mark scheme what they did to get dv/dt. The final answer should be 3.
    Thanks
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    (Original post by dont)
    Can someone help me with this question?

    v = 20000te^(-t/3)
    Use this model and calculus to find the value of t when dv/dt = 0

    I can't work out from the mark scheme what they did to get dv/dt. The final answer should be 3.
    Thanks
    Product rule?
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    \displaystyle \frac{\mathrm{d}V}{\mathrm{d}t} = 20000\left[e^{-t/3} \frac{\mathrm{d}}{\mathrm{d}t}\{  t\} +t \frac{ \mathrm{d}}{ \mathrm{d} t}\{e^{-t/3}\}\right]
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    (Original post by dont)
    Can someone help me with this question?

    v = 20000te^(-t/3)
    Use this model and calculus to find the value of t when dv/dt = 0

    I can't work out from the mark scheme what they did to get dv/dt. The final answer should be 3.
    Thanks
    As mentioned above, you would use the product rule for this
    You should get dv/dt = 2000e^(-t/3) - ((2000/3)t^2)e^(-t/3)
    Spoiler:
    Show
    Sorry for the messy layout I'm on mobile

    You are correct that this would need to equal 0, you could try to factorise this to make it a bit nicer which would help you solve it too.
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    (Original post by KaylaB)
    As mentioned above, you would use the product rule for this
    You should get dv/dt = 2000e^(-t/3) - ((2000/3)t^2)e^(-t/3)
    Spoiler:
    Show
    Sorry for the messy layout I'm on mobile
    You are correct that this would need to equal 0, you could try to factorise this to make it a bit nicer which would help you solve it too.
    Ah thanks. Could you possible go through the steps you did to get dv/dt as I struggle with the product rule.
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    (Original post by dont)
    Ah thanks. Could you possible go through the steps you did to get dv/dt as I struggle with the product rule.
    You split the functions in to two parts, usually shown as f(x) = uv where you're able to diffrentiate u and v individually
    So f'(x) = (du/dx)v + (dv/dx)u
    So basically you diffrentiate one and multiply it by the original of the other. Then do vice versa and add them
    So if f(x) = 4xe^2x = (4x)(e^2x) (So u=4x and v=e^2x)
    du/dx = 4
    dv/dx = 2e^2x
    So f'(x) = (4)(e^2x) + (2e^2x)(4x) = 4e^2x + 8xe^2x
    = (4e^2x)(1+2x) which is just simplified
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    (Original post by KaylaB)
    You split the functions in to two parts, usually shown as f(x) = uv where you're able to diffrentiate u and v individually
    So f'(x) = (du/dx)v + (dv/dx)u
    So basically you diffrentiate one and multiply it by the original of the other. Then do vice versa and add them
    So if f(x) = 4xe^2x = (4x)(e^2x) (So u=4x and v=e^2x)
    du/dx = 4
    dv/dx = 2e^2x
    So f'(x) = (4)(e^2x) + (2e^2x)(4x) = 4e^2x + 8xe^2x
    = (4e^2x)(1+2x) which is just simplified
    I get it now. Thanks that really helped
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    (Original post by dont)
    I get it now. Thanks that really helped
    No problem
 
 
 

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