# 1 Math 171 Exam I Answers February 14, 2002

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Math 171
February 14, 2002
√
−
→
→
−
−
1. →
a =&lt; 4, 2 &gt; so a vector that is orthogonal is b =&lt; −2, 4 &gt;. Now | b | = 20
3 −
−6
12
→
Answer: √ b = √ , √
20
20
20
→
−
2. The vector that represents the path of the object is d =&lt; 3, 1 &gt;. Since
→
→ −
−
Work = F &middot; d = 15 + 6 = 21J
3. Remember that j =&lt; 0, 1 &gt;. so the vector projectin of &lt; 5, 9 &gt; onto j is &lt; 0, 9 &gt;.
4. Solve x = 1+cos(t) and y = 1+sin2 (t) for cos(t) and sin2 (t) and subsitute that into the equation
sin2 (t) + cos2 (t) = 1.
Answer: (y − 1) + (x − 1)2 = 1 or y = 1 + 2x − x2
5. (a) r(2) = 6i − 2j
(b) If the particle goes through the point (4, 8), then there is some value of t such that
8 = t − 4 which implies that t = 12. Since 122 + 12 6= 4, the particle can not go through
the point (4, 8).
f (x + h) − f (x)
. Look at the class notes and the
h→0
h
material posted on the web to find the solution.
6. The definition of the derivative is f ′ (x) = lim
7. If this function is to be continuous at x=1 and x=2, then we need:
lim f (x) = lim f (x) = f (1) and lim f (x) = lim f (x) = f (2)
x→1+
x→1−
x→2+
x→2−
This means that 2 = c + d and 4c + d = 8. Now solving for c and d gives c = 2 and d = 0.
8. (30 points) Find the exact values of the following limits:
(a) lim
t→1
*
t2 + t − 2 3
, t − 4t
t2 − t − 2
+
=
*
t2 + t − 2
lim 2
, lim t3 − 4t
t→1 t − t − 2 t→1
+
= h0, −3i
x+1
= +∞ notice that x = 3 is a vertical asymptote. so plug in values of x
− 3)
x→3
slightly above 3 to see which direction the function goes.
(b) lim+
x2 (x
(x − 4)(x + 2)
x+2
6
x2 − 2x − 8
= lim
= lim
=
= −3
x→4 (x − 4)(x − 6)
x→4 x − 6
x→4 x2 − 10x + 24
−2
(c) lim
1
(d) lim cos(x) = 0 since the cos(x) function is always between -1 and 1 and the function
x→∞ x
gets closer to zero as x → ∞.
(e)
lim (x +
x→−∞
p
x2
+ 4x + 1) = lim (x +
x→−∞
p
x2
√
(x − x2 + 4x + 1)
√
+ 4x + 1) ∗
=
(x − x2 + 4x + 1)
−4x − 1
x2 − (x2 + 4x + 1)
√
√
= lim
∗
x→−∞ (x − x2 + 4x + 1)
x→−∞ (x − x2 + 4x + 1)
lim
−4 − x1
−4 − x1
√
q
=
lim
−1
x→−∞ (1 − √
x2 + 4x + 1) x→−∞ (1 + 1 + 4 +
2
lim
x
x
1
)
x2
1
x
1
x
−4 − x1
√
=
x→−∞ (1 − 1 x2 + 4x + 1)
x
= lim
=
1
x
−4
√ = −2
1+ 1
9. (a) lim f (x) = 2
x→9
(b)
lim f (x) = 4
x→−3+
(c) lim f (x) = −∞
x→7−
(d) For what values of x is the function not differentiable? x = −3, 3, 7, 9
(e) For what values of x is the function not continuous? x = −3, 7, 9
(f) Approximate the value of f ′ (4). If not possible, explain why.
Draw the tangent line in the graph and then approximate the slope.
My answer is f ′ (4) ≈ 1.35
2
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