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    http://www.ocr.org.uk/Images/59547-q...athematics.pdf

    Question 9ii)

    For the constant c I get ln140 in the mark scheme it says -ln140 I can't seem to figure out why.

    My working out :

    1 / ( 160 - theta) d theta = k dt

    ln ( 160 - theta) = kt + c

    When t = 0 theta = 20

    ln140 = 0 + c
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    (Original post by IShouldBeRevising_)
    http://www.ocr.org.uk/Images/61388-q...hematics-2.pdf

    Question 9ii)

    For the constant c I get ln140 in the mark scheme it says -ln140 I can't seem to figure out why.

    My working out :

    1 / ( 160 - theta) d theta = k dt

    ln ( 10 - theta) = kt + c

    When t = 0 theta = 20

    ln140 = 0 + c
    you missed a minus sign when you integrated the LHS!
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    (Original post by davros)
    you missed a minus sign when you integrated the LHS!


    I don't understand
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    (Original post by IShouldBeRevising_)
    http://www.ocr.org.uk/Images/59547-q...athematics.pdf

    Question 9ii)

    For the constant c I get ln140 in the mark scheme it says -ln140 I can't seem to figure out why.

    My working out :

    1 / ( 160 - theta) d theta = k dt

    ln ( 160 - theta) = kt + c

    When t = 0 theta = 20

    ln140 = 0 + c
    Temperature is increasing so:
    \frac{d\theta}{dt} = k(160-\theta)
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    (Original post by joostan)
    Temperature is decreasing so:
    \frac{d\theta}{dt} = -kt
    You can say k, but then you must remember that k is a negative constant.
    The question says increasing ?
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    (Original post by IShouldBeRevising_)
    The question says increasing ?
    ah, misread the q COMPLETELY! poor show see above.
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    (Original post by IShouldBeRevising_)
    The question says increasing ?
    Allow me to try again.
    \int \frac{(f'x)}{f(x)}\ dx = ln|x|
    what does 160-theta differentiate to?
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    (Original post by joostan)
    Allow me to try again.
    \int \frac{(f'x)}{f(x)}\ dx = ln|x|
    what does 160-theta differentiate to?
    -1 ?
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    (Original post by IShouldBeRevising_)
    -1 ?
    So as davros says - where is the minus on the LHS?
    f'(x)/f(x) = -1/(160-theta)
    So the integral should read -[-1/(160-theta)]
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    (Original post by joostan)
    So as davros says - where is the minus on the LHS?
    f'(x)/f(x) = -1/(160-theta)
    So the integral should read -[-1/(160-theta)]
    But originally the lhs = 1/160 - thetA

    d/dx(1/ 160 - theta ) = -1 however the original lhs is 1 not -1?
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    (Original post by IShouldBeRevising_)
    But originally the lhs = 1/160 - thetA

    d/dx(1/ 160 - theta ) = -1 however the original lhs is 1 not -1?
    when you integrate 1/(160 - theta) you should get -ln|160 - theta|
 
 
 
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