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Unformatted text preview: Solutions to Quiz 8A
www.math.uﬂ.edu/˜harringt
March 20, 2008
1. A particle moves according to a law of motion s = f (t) = 2t3 − 15t2 +
36t, for t ≥ 0, where t is measured in seconds and s is in feet.
(a) Find the velocity at time t.
v (t) = ds/st = f (t) = 6t2 − 30t + 36
(b) When is the particle at rest?
When v (t) = 0. So,
0 = 6t2 − 30t + 36 = 6(t2 − 5t + 6) = 6(t − 2)(t − 3)
Hence, the particle is a rest at 2 seconds and 3 seconds.
(c) When is the particle moving in the positive direction?
When v (t) > 0. Recall that t ≥ 0. Here I will use a sign chart on
the factors found in part (b).
6
t2
t3
v(t) +
+ ++++
0+++
0+
00+
2
3 So the particle is increasing on [0, 2) ∪ (3, ∞).
2. Find the diﬀerential dy and evaluate dy for the given values of x and
dx 1 (a) y = ex/2 , x = 0, dx = 1
10 1
dy = ex/2 dx
2
After substition we have:
1
11
dy = e0 =
2 10
20
(b) y = sin(x), x = π , dx =
3 1
100 dy = cos(x)dx
After substition we have:
dy = cos π
1
1
1
1
∗
=∗
=
3
100
2 100
200 2 ...
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 Fall '08
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 Math, Calculus, Geometry

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