Calculus II |
Quiz #4
|
March 6, 2002
|
Name____________________ |
R. Hammack
|
Score ______
|
(1) Evaluate the definite integrals. Simplify your answer as much as possible
(a) ![[Graphics:Images/quiz4sol_gr_1.gif]](Images/quiz4sol_gr_1.gif)
![[Graphics:Images/quiz4sol_gr_2.gif]](Images/quiz4sol_gr_2.gif)
=![[Graphics:Images/quiz4sol_gr_4.gif]](Images/quiz4sol_gr_4.gif)
=![[Graphics:Images/quiz4sol_gr_6.gif]](Images/quiz4sol_gr_6.gif)
u=πx
du=πdx
(b) ![[Graphics:Images/quiz4sol_gr_7.gif]](Images/quiz4sol_gr_7.gif)
![[Graphics:Images/quiz4sol_gr_8.gif]](Images/quiz4sol_gr_8.gif)
=ln(e+5)
- ln(6) = ln((e+5)/6)
u = ![[Graphics:Images/quiz4sol_gr_10.gif]](Images/quiz4sol_gr_10.gif)
du =
dx
(2) Find the derivative of the function F(x)
= ![[Graphics:Images/quiz4sol_gr_12.gif]](Images/quiz4sol_gr_12.gif)
Using the chain rule and Part 2 of the Fundamental Theorem of Calculus, we get
F '(x) = sin(
-(x+ln(x))
)(1 + 1/x)
(3) A freight train, moving with constant acceleration, travels 20 miles
in half an hour. At the beginning of the half-hour period, it has a
velocity of 10 miles per hour. What is its velocity at the end of the
half-hour period?
Let's set the clock so that t=0 represents the beginning of the half-hour
period and t = 1/2 represents the end.
Represent the constant acceleration as a, so the velocity is v(t)
=
= at +C.
Since it's given that 10 = v(0) = a(0) + C, we get C
= 10.
Therefore the velocity is v(t) = at
+10. If we can just find a, the answer to the question will
be v(1/2).
To use the information about position (i.e. the fact that the train went 20 miles),
we must now construct the position function.
Notice that s(t) =
=
=
,
and from 0 = s(0) =
we obtain K=0.
Therefore, position is s(t) =![[Graphics:Images/quiz4sol_gr_19.gif]](Images/quiz4sol_gr_19.gif)
Since the train went 20 miles in half an hour, we know that 20 = s(1/2)
= ![[Graphics:Images/quiz4sol_gr_20.gif]](Images/quiz4sol_gr_20.gif)
which becomes 20 = a/8 +5. Solving for a, we get a = 120.
Therefore, we can finally write velocity as v(t)
= 120t +10.
Here is the answer to the question. The velocity at the end of the half-hour
period is
v(1/2) = 120(1/2)+10 = 70 miles per hour.