Download Math 3: Calculus, Muetzel Winter 2016 - Solution 20

Math 3: Calculus, Muetzel
Winter 2016 - Solution 20
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problem 1.
(10 points)
antiderivative
Find the most general antiderivatives of the following functions.
1.) f (x) = π .
The most general antiderivative of f (x) is F (x) = π · x + C , where C is a constant.
2.) g(x) = x98 .
The most general antiderivative of g(x) is
G(x) =
−9
7x7
−9
7x7
+ C1
+ C2
if x < 0
if x > 0
,where C1 and C2 are constants.
√
3.) h(x) = 2 x + 5 cos(x + 1) = 2x1/2 + 5 cos(x + 1).
The most general antiderivative of h(x) is H(x) = 4/3 · x3/2 + 5 sin(x + 1) + C , where C
is a constant.
4.) f˜(mx + b), given that F̃ 0 (x) = f˜(x).
The most general antiderivative of f˜(x) is
problem 2.
(10 points)
F̃ (mx+b)
m
+ C , where C is a constant.
You start your car at the trac light. You accelerate linearly. After
10 seconds your speed is 30 ft/s (about 20 mph). Find your distance from the trac light and
your acceleration after 10 seconds.
This is the example from Lecture 12. The general idea is that the derivative of a polynomial is a polynomial. When dierentiating the degree of the polynomial goes down by one.
1.) a(t) = m · t + b: acceleration of the car. We have that a(0) = 0, hence a(t) = m · t.
2.) v(t): speed of the car. We have that
v 0 (t) = a(t) = m · t, hence v(t) =
m 2
t + c.
2
As v(0) = 0, v(10) = 30, we get that v(t) = 0.3 · t2 and m = 0.6, hence a(t) = 0.6 · t.
2.) d(t): distance of the car from the trac light. We have that
d0 (t) = v(t), hence d(t) = 0.1 · t3 + k.
As d(0) = 0, we have that d(t) = 0.1 · t3 . Evaluating d(t) and a(t) at t = 10, we get that
d(10) = 100 ft and a(10) = 60f t/s2 .
Math 3: Calculus, Muetzel
Winter 2016 - Solution 20
problem 3.
Let f (x) be dened as below. Draw f (x) and graph an antiderivative of f (x).
f (x) =
if x ∈ [−3, 1]
.
if x ∈ (1, 3]
|x|
1
You can do this!
An antiderivative F (x) of f (x) is
F (x) =

2

 − x2
x2
2

 x−
1
2
if x ∈ [−3, 0]
if x ∈ (0, 1] .
if x ∈ (1, 3]