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Thread starter 14 years ago
#1
I am only posting these questions because my calculus is not up to C3/C4 yet. These are in ex 1B.

9) A particle P moves along the positive x-axis and its acceleration is (e^2x)ms^-2 when its displacement from O is x metres. Given that initially, when t = 0 and x = ln u, obtain a) velocity of P as a function of x, b) as a function of t.

11) A particle P moves on the positive x-axis and its acceleration when it has a displacement of x metres from O, is (4sinx)ms^-2 in the direction Ox. Given that when x=0 the velocity of P is 6ms^-1 in the direction Ox, find the maximum speed of P.

12) A particle P moves on the positive x-axis and its acceleration when it has a displacement of x metres from O, is (e^2x)ms^-2. Given that initially when t = 0, the speed of P is 1ms^-1 in the direction Ox and x=0 obtain x as a function of t and show that x tends towards infinity as t tends towards 1.
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14 years ago
#2
Derivative of e^f(x) = f'(x)e^f(x).
Integral of e^(ax+b) = (1/a)e^(ax+b).

Derivative of : sin(f(x)) = f'(x)cos(f(x)), cos(f(x)) = -f'(x)sin(f(x))
Integral of: sin(ax+b) = -(1/a)cos(ax+b), cos(ax+b) = (1/a)sin(ax+b)
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Thread starter 14 years ago
#3
(Original post by SsEe)
Derivative of e^f(x) = f'(x)e^f(x).
Integral of e^(ax+b) = (1/a)e^(ax+b).

Derivative of : sin(f(x)) = f'(x)cos(f(x)), cos(f(x)) = -f'(x)sin(f(x))
Integral of: sin(ax+b) = -(1/a)cos(ax+b), cos(ax+b) = (1/a)sin(ax+b)
I have an expression
t = ln(e^2x - 1)

How do I make x the subject?
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14 years ago
#4
(Original post by Widowmaker)
I have an expression
t = ln(e^2x - 1)

How do I make x the subject?
t = ln(e^2x - 1)
---> e^t = e^2x - 1
---> e^2x = e^t + 1
---> 2x = ln(e^t + 1)
---> x = ln[Sqrt.(e^t + 1)]
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