Download MATH 114 W09 Test 2 Solutions 1 1. Find the derivatives of the

MATH 114 W09 Test 2 Solutions
1
1. Find the derivatives of the following functions:
a) f (x) = (x2 + sin x + cos x)(x5 + 4x3 + 7x).
Solution: Use the product rule:
f 0 (x) = (2x + cos x − sin x)(x5 + 4x3 + 7x) + (x2 + sin x + cos x)(5x4 + 12x2 + 7).
ex + x2
.
(ln x) + 1
Solution: Use the quotient rule:
b) g(x) =
f 0 (x) =
(ex + 2x)((ln x) + 1) − (ex + x2 ) x1
.
((ln x) + 1)2
2. Differentiate using the chain rule:
√
a) f (x) = ln(1 + x2 + x).
Solution:
1
1
√
f 0 (x) =
· √
· (2x + 1).
2
1 + x + x 2 x2 + x
b) f (x) = sin(tan(1 + x)).
Solution: f 0 (x) = cos(tan(1 + x)) · sec2 (1 + x).
3. A particle moves on a vertical line so that its vertical distance y at time t is given by
y = t3 − 12t + 3, t ≥ 0.
a) Find the velocity and acceleration functions.
Solution: v(t) = dy
= 3t2 − 12. a(t) = dv
= 6t.
dt
dt
b) When is it moving upward and when is it moving downward?
Solution: It is moving upward when v(t) > 0, or when 3t2 − 12 > . This happens
when t2 > 4, or t > 2.
The particle is moving downward when v(t) < 0. The is happens when 0 ≤ t < 2.
c) When is the particle speeding up? When is it slowing down?
Solution: The particle is speeding up when a(t) > 0, or when 6t > 0. The happens
when t > 0. In other words, the particle is always speeding up, and never slowing
down.
(6pts)
4. Use implicit differentiation to find
dy
for
dx
a) xy 4 + x2 y = x + 3y
Solution: Take the derivative of both sides of the equation:
y 4 + 4xy 3 y 0 + 2xy + x2 y 0 = 1 + 3y 0 ,
(4xy 3 + x2 − 3)y 0 = 1 − 2xy − y 4 ,
1 − 2xy − y 4
.
y0 =
4xy 3 + x2 − 3
MATH 114 W09 Test 2 Solutions
b)
2
√
xy = 1 + x2 y.
Solution: Take the derivative of both sides:
1
√ · (y + xy 0 ) = 2xy + x2 y 0
2 xy
y
x
√ + √ · y 0 = 2xy + x2 y 0 ,
2 xy 2 xy
x
y
( √ − x2 )y 0 = 2xy − √ ,
2 xy
2 xy
2xy − 2√yxy
0
y =
√x − x2
2 xy
5. Use linear approximation to the function f (x) =
√
x at a = 4 to approximate
√
4.05.
Solution: For x near a we have the linear approximation f (x) ' f (a) + f 0 (a)(x − a).
We have that f 0 (x) = 2√1 x , and thus
√
x'
√
1
a + √ (x − a).
2 a
When x = 4.05 and a = 4, we get
√
4.05 '
√
1
1
4 + √ (4.05 − 4) = 2 + · 0.05 = 2.0125.
4
2 4
6. Find the absolute maximum value for f (x) = x4 − 2x2 + 3 on the interval [−2, 3].
Solution: First find the critical values: f 0 (x) = 4x3 − 4x. Set f 0 (x) = 0, and solve. We
get 4x3 − 4x = 0, x(x2 − 1) = 0. So there are 3 solutions: x = 0, ±1. To check for max’s
and min’s, take the second derivative: f 00 (x) = 12x2 − 4. We have f 00 (0) = −4 < 0,
and so x = 0 is a local max. We have f 00 (1) = 8 > 0, and f 00 (−1) = 8 > 0, so both
x = 1, and x = −1 are local mins. To determine the absolute maximum, we need to
compare f (0) = 3 with the function values at the endpoints. We have f (−2) = 11,
and f (3) = 66. So f has an absolute maximum value of 66.
7. A car travelling north at 40 km/h and a truck travelling east at 30 km/h leave an
intersection at the same time. At what rate will the distance between them be changing
3 hours later?
Solution: Let x(t) be the distance the truck travels in time t and let y(t) be the distance
the car travels in time t. Let z(t) be the distance between the car and truck at time
t. We are given x0 (t) = 30 km/h and y 0 (t) = 40 km/h. Furthermore, by Pythagoras’
theorem, x(t)2 +y(t)2 = z(t)2 . Differentiating both sides of the equation (using the chain
rule), we obtain 2x(t)x0 (t) + 2y(t)y 0 (t) = 2z(t)z 0 (t). So x(t)x0 (t) +p
y(t)y 0 (t) = z(t)z 0 (t).
After 3 hours, x(3) = 90 km and y(3) = 120 km. Then z(3) = 90)2 + 1202 = 150
km. Substituting into the equation, 90 · 30 + 120 · 40 = 150 · z 0 (3). Solving for z 0 (3),
we obtain, z 0 (3) = 50 km/h.
MATH 114 W09 Test 2 Solutions
3
Bonus The angle of elevation of the Sun is decreasing at a rate of 0.25 rad/hour. How fast
is the shadow cast by a 400-metre-tall building increasing when the angle of elevation of the
π
Sun is ?
6
Solution: Let θ(t) be the angle of the sun at time t. Let x(t) be the length of the shadow at
400
. Taking the derivative of
time t (measured from the base of the building). Then tan θ = x(t)
both sides, we get
400
sec2 (θ)θ0 (t) = − 2 x0 (t).
x
π
0
We have θ (t) = −0.25 rad/hour. When θ = 6 , we have sec π6 = √23 , and tan π6 = √13 . Then
√ . Substituting these values into the above equation, we get
x(t) = 400
3
4
400
(−0.25) = − 4002 x0 (t).
3
3
Solving for x0 , we get x0 =
400
9
m/h.