A particle is moving so that the position at time t is given by:
r(t) = e^t(cost i - sint j)
Find the velocity vector at time t:
find the acceleration vector at time t:
find the speed at time t:
i think i need to find the first and second derivatives for velocity and acceleration, not sure about speed?
i wasn't sure on the derivatives of a natural log.
thanks for any help.
might need some help fixing my differentiation skills..
for the first derivative my guess would be to use the chain rule to differentiate?
so the first derivative is: -1/x(cost i - sin j)(sint i - cost j)
and for the 2nd derivative i guessed to use the chain and product rule in combination: so i got something like:- 1/x^2(cost i - sin j)(sint i - cost j)[(sint i - cost j)(sint i - cost j) + (cost i - sint j)(cost i + sint i)]
how am i doing.
thanks.
What is x??
I think you are making this too hard. Given h(x) = f(x)*g(x)
h'(x) = f'(x)*g(x) + f(x)*g'(x)
So:
r(t) = e^t*(cos(t)i - sin(t)j)
r'(t) = e^t*(cos(t)i - sin(t)j) + e^t*(-sin(t)i - cos(t)j)
r'(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j)
r''(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j) + e^t*([-sin(t) - cos(t)]i + [-cos(t) + sin(t)]j)
r''(t) = e^t*(-2sin(t)i + -2cos(t)j)
r''(t) = -2e^t*(sin(t)i + cos(t)j)
For the speed:
r'(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j)
So
|r'(t)| = sqrt{e^(2t)[cos(t) - sin(t)]^2 + e^(2t)[-sin(t) - cos(t)]^2}
|r'(t)| = e^t*sqrt{cos^2(t) + sin^2(t) - 2sin(t)cos(t) + sin^2(t) + cos^2(t) + 2sin(t)cos(t)}
|r'(t)| = e^t*sqrt{1 + 1}
|r'(t)| = e^t*sqrt{2}
-Dan