NAME:
MATH 151
October 15, 2014
QUIZ 5
• Calculators are NOT allowed!
• Show all your work and indicate your final answer clearly. You will be graded not merely
on the final answer, but also on the work leading up to it.
1. (3 points) Find y 0 by implicit differentiation:
10 12
+ 2 = 2014
x
y
Solution: Rewriting the equation gives
10 12
2014 =
+ 2 = 10x−1 + 12y −2 .
x
y
Differentiating across using the “power rule” gives
−10 24
0 = −10x−2 − 24y −3 y 0 = 2 − 3 y 0
x
y
24
−10
⇔ 3 y0 = 2
y
x
−5y 3
−10y 3
=
.
⇔ y0 =
24x2
12x2
2. (3 points) Find the slope of the tangent line at x = π for the function f (x) = sin(cos(3x)).
Solution:We use the chain rule:
f 0 (x) = cos(cos(3x)) · (− sin(3x)) · 3
and so the slop of the tangent line of f at x = π is
f 0 (π) = cos(cos(3π)) · (− sin(3π)) · 3 = cos(1) · 0 · 3 = 0.
NAME:
MATH 151
October 15, 2014
3. (3 points) The vector function r(t) represents the position of a particle at time t. Find the
velocity and speed of the particle at time t = π4 where
√
r(t) = ( t)i + (cos t)j.
Solution:The velocity of the particle is given by r0 (t):
1
r0 (t) = √ i + (− sin(t))j
2 t
so the veclocity of the particle at time t = π4 is
1
π
0 π
r
= p π i + − sin
j
4
4
2 4
1
1
= √π i + − √ j
2
2 2
1
1
= √ i − √ j.
π
2
Speed s is the magnitude of velocity so
s
2 2 r
0 π π
−1
1 1
1
=
√
s
+ .
=
r
+ √
=
4
4
π 2
π
2