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Expert: - 5/6/2011


Question
I need help on this quickly please!


A particle moves along a line so that its velocity at time t is v(t) = t^2 - 2 (measured) in meters per second.

a.) Find the displacement of the particle during the time period 0 ≤ t ≤ 2. (i.e. where is the particle located after 2 seconds?)

b.) Find the distance traveled during this time period.



Find the derivative of each function.

Given F(x)=∫ln(tē+5) dt; a=0, b=x.

i. Find F'(x).

ii. If H(x)=∫ln(tē+5)dt, [a=0, b=sinx] express H(x) as a composition of functions, one of which should be F. Compute H'(x).

iii. If G(x)=∫ln(tē+5)dt, [a=cosx, b=sinx], compute G'(x).

Please show your work and steps. Thanks in advance!

Answer
1
The distance is the integral of velocity.
That gives t^3/3 - 2t + C.
At t=0, the position is C.
At t=2, the position is 8/3 - 4 + C = 8/3 - 12/3 + C = -4/3 + C.
The difference is 4/3, and that is the distance travelled backwards,
so it is at -4/3 from where it was.


2
i.If F(x) is an integral, then F'(x) is what is being integrated.
Thus, F(x) = ln(tē+5).

ii. It can be seen that H(x) = F(sinx) - F(0).

iii. The derivative of an integral is the function, so G'(x) = ln(tē+5).
For a=cosx and b=sinx, it would be ln(sinēx+5) - ln(cosēx + 5).
The subtraction of ln() means to divide what is inside, so it is ln[(sinēx+5)/(cosē(x)+5)].

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