Download Math 1A, Spring 2008 Midterm 1 Solutions - Prof. J. A. Wilkening and more Exams Calculus in PDF only on Docsity!
Math 1A, Spring 2008, Wilkening
Sample Midterm 1
- (2 points) What is the domain of the function f (x) =
x − 2 +
7 − x
x − 5
Answer:
[2, 5) ∪ (5, 7]
x must be ≥ 2 for
x − 2 to be defined. It must be ≤ 7 for
7 − x to be defined,
and it must 6 = 5 so that x − 5 6 = 0
- (3 points) Compute the following derivative:
d
dθ
θ=
tan θ cos θ
Answer:
d
dθ
θ=
tan θ cos θ =
d
dθ
θ=
sin θ
cos θ
cos θ =
d
dθ
θ=
sin θ = cos 0 = 1
- (5 points) Find the equation of the tangent line to the curve y = 1/(1 + x^2 ) at the
point
1 10
Answer:
dy
dx
− 2 x
(1 + x^2 )^2
so at
1 10
, the slope of the tangent line is − 2 ∗ 3 (1+3^2 )^2 =^
− 3
Putting the line into point-slope format gives y − 1 10 =^
− 3 50 (x^ −^ 3) or^ y^ =^
− 3 50 x^ +^
7 25
- (5 points) Compute lim x→∞
(x − 1)(2x − 3)
(4x + 1)(7x + 1)
Answer:
lim x→∞
(x − 1)(2x − 3)
(4x + 1)(7x + 1)
= lim x→∞
(1 − 1 /x)(2 − 3 /x)
(4 + 1/x)(7 + 1/x)
- (5 points) Prove that f (x) = 1 − x^5 has a fixed point (i.e. there is a number c such
that f (c) = c).
Answer:
Let g(x) = f (x) − x = 1 − x^5 − x
Then, g(0) = 1 > 0 and g(1) = − 1 < 0, so by the Intermediate Value Theorem, there
is some c in (0, 1) such that g(c) = 0, and therefore, for that same c, f (c) = c
- (5 points) Evaluate the limit
lim θ→ 0 −
sin θ √ 1 − cos θ
.
Answer:
Note first that when θ < 0 but θ is close to 0, sin θ < 0
so sin θ = −
sin 2 θ = −
1 − cos^2 θ, thus we get
lim θ→ 0 −
sin θ √ 1 − cos θ
= lim θ→ 0 −
1 − cos^2 θ √ 1 − cos θ
= lim θ→ 0 −^
1 − cos^2 θ
1 − cos θ
= lim θ→ 0 −^
1 + cos θ = −
- (5 points) Use the δ-ε definition of the limit to prove one of the following:
(a) lim x→ 2
1 /x = 1/2.
Answer:
Let ε > 0. Choose δ =min(2ε, 1). Then, if 0 < |x − 2 | < δ, then
| 1 /x − 1 / 2 | = |(2 − x)/ 2 x| = |x − 2 |/| 2 x| < δ/| 2 x| <^1 δ/ 2 ≤ ε
(^1) : note that δ ≤ 1, so if 0 < |x − 2 | < δ then x > 1 so δ/| 2 x| < δ/ 2
(b) Suppose lim x→ 0
f (x) = L. Define g(x) = f (−x). Then lim x→ 0
g(x) = L.
Answer:
Let ε > 0. Choose δ so that if 0 < |x − 0 | < δ then |f (x) − L| < ε
(such a δ exists because lim x→ 0
f (x) = L)
Then, for the chosen δ, if 0 < |x − 0 | < δ then 0 < |(−x) − 0 | < δ so
|g(x) − L| = |f (−x) − L| < ε
- (5 points) Evaluate the limit
lim x→−∞
x^2 + x + 1 −
x^2 + 1
Answer:
lim x→−∞
x^2 + x + 1 −
x^2 + 1
= lim x→−∞
x^2 + x + 1 −
x^2 + 1
x^2 + x + 1 +
x^2 + 1
x^2 + x + 1 +
x^2 + 1
lim x→−∞
(x^2 + x + 1) − (x^2 + 1) (√ x^2 + x + 1 +
x^2 + 1
) = lim x→−∞
x (√ x^2 + x + 1 +
x^2 + 1
lim x→−∞
x/
x^2 (√ 1 + 1/x + 1/x^2 +
1 + 1/x^2
- (5 points) A stone is dropped from height H on a planet with gravitational constant
g. The equation of motion of the stone is y(t) = H− 12 gt^2. Show that the instantaneous
velocity of the stone when it hits the ground is twice the average velocity during its
fall.
Answer:
Let T be the time that the stone hits the ground. Then 0 = y(T ) = H − 1 2 gT^
so H = 12 gT 2 and thus T =
2 H/g
The average velocity, vavg is (the total distance traveled)/(the total time traveled),
vavg = −H/T = −H/(
2 H/g) = −
Hg/ 2
(the velocity is negative because the stone traveled down)
The velocity at time t is given by y ′ (t) = −gt, therefore the final velocity vf in is y ′ (T )
vf in = −gT = −g
2 H/g = −
2 Hg = − 2
Hg/2 = 2vavg
