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Differential Equation Velocity Time

The (x,v) equation for a particle is v = 10 - 0.1x^2 for 0 < x < 10. Find the (t,x) equation, given that x = 0 when t = 0.

Can you please help me figure out the error in my working?

Unparseable latex formula:

v = 10 - 0.1x^2 \\[br]\dfrac{\mathrm{d} x}{\mathrm{d} t} = 10 - 0.1x^2\\[br]\dfrac{\mathrm{d} t}{\mathrm{d} x} = \dfrac{1}{10 - 0.1x^2} = \dfrac{10}{100 - x^2}\\[br]= \dfrac{10}{(10 - x)(10 + x)} = \dfrac{1}{2} \left( \dfrac{1}{10 - x} + \dfrac{1}{10 + x} \right)


Unparseable latex formula:

t = \dfrac{1}{2} \int \dfrac{1}{10 - x} + \dfrac{1}{10 + x} \, \mathrm{d} x\\[br]t = \dfrac{1}{2}\left( \ln(10-x) + \ln(10 + x) \right ) + c



Since v = 0 when t = 0,

Unparseable latex formula:

0 = \ln(10) + c\\[br]c = -\ln(10)\\[br]t = \dfrac{1}{2}\left( \ln(10-x) + \ln(10 + x) \right ) - \ln(10)\\[br]\dfrac{1}{2}\left( \ln(10-x) + \ln(10 + x) \right ) = t + \ln(10)\\[br]\ln(10-x) + \ln(10 + x) = ln(100 - x^2) = 2(t + \ln(10))\\[br]100 - x^2 = \exp(2(t + \ln(10))) = \exp(2t + 2\ln(10)) = \exp(2t + \ln(100))\\[br]100 - x^2 = e^{2t + \ln(100)} = e^{2t}e^{\ln(100)} = 100e^{2t}\\[br]x = \sqrt{100 - 100e^{2t}} = 10\sqrt{1-e^{2t}}



But my textbook says v=10(e2t1e2t+1)v = 10\left(\dfrac{e^{2t} - 1}{e^{2t} + 1}\right).
(edited 4 years ago)
Reply 1
Original post by esrever
The (x,v) equation for a particle is v = 10 - 0.1x^2 for 0 < x < 10. Find the (t,x) equation, given that x = 0 when t = 0.

Can you please help me figure out the error in my working?

Unparseable latex formula:

v = 10 - 0.1x^2 \\[br]\dfrac{\mathrm{d} x}{\mathrm{d} t} = 10 - 0.1x^2\\[br]\dfrac{\mathrm{d} t}{\mathrm{d} x} = \dfrac{1}{10 - 0.1x^2} = \dfrac{10}{100 - x^2}\\[br]= \dfrac{10}{(10 - x)(10 x)} = \dfrac{1}{2} \left( \dfrac{1}{10 - x} \dfrac{1}{10 x} \right)


Unparseable latex formula:

t = \dfrac{1}{2} \int \dfrac{1}{10 - x} \dfrac{1}{10 x} \, \mathrm{d} x\\[br]t = \dfrac{1}{2}\left( \ln(10-x) \ln(10 x) \right ) c



Since v = 0 when t = 0,

Unparseable latex formula:

0 = \ln(10) c\\[br]c = -\ln(10)\\[br]t = \dfrac{1}{2}\left( \ln(10-x) \ln(10 x) \right ) - \ln(10)\\[br]\dfrac{1}{2}\left( \ln(10-x) \ln(10 x) \right ) = t \ln(10)\\[br]\ln(10-x) \ln(10 x) = ln(100 - x^2) = 2(t \ln(10))\\[br]100 - x^2 = \exp(2(t \ln(10))) = \exp(2t 2\ln(10)) = \exp(2t \ln(100))\\[br]100 - x^2 = e^{2t \ln(100)} = e^{2t}e^{\ln(100)} = 100e^{2t}\\[br]x = \sqrt{100 - 100e^{2t}} = 10\sqrt{1-e^{2t}}



But my textbook says v=10(e2t1e2t1)v = 10\left(\dfrac{e^{2t} - 1}{e^{2t} 1}\right).


Update: I found the answer. Integral of 1/(10-x) is -ln(10-x).
(edited 4 years ago)

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