Homework 10 - SOLUTIONS
(100 points)
1. (10 points) Find the derivative of these functions by applying the Power Rule:
a) f(x)= x4
Solution:
f’(x)= 4x3
b) y= =x-2
Solution:
f’(x) = -2x-3
2. (10 points) Find the derivative of these functions by applying the Product Rule:
a) y=(3x-2x2)(5+4x)
Solution:
y’ = (3x-2x2) (4)+ (5+4x)(3-4x)
= (12x -8x2) + 15 -20x +12x -16x2
= 4x – 24x2 +15
b) f(x) =
+1)(x-1)
Solution:
Rewriting f(x)
f(x) =
+1)(x-1)
f’(x) =
+1)(1) + (x-1)(-x-2)
f’(x) = + 1 Finding a common denominator
f’(x) =
f’(x) =
3. (10 points) Find the derivative of these functions by applying the Quotient Rule:
a) y=
Solution:
b) y=
Solution:
y=
Multiply numerator and denominator by x
=
=
=
(
)
(
=
)
=
4. (10 points) Find the derivative of these functions by applying the Chain Rule:
a)
y=(x2 + 1)3
Solution:
(2x) = 6x(x2 + 1)2
b) f(x)=(3x-2x2)3
Solution:
f’(x)=3(3x-2x2)2 (3-4x)
=(9-12x) (3x-2x2)2
5. (30 points) A model rocket if fired from the roof of a 50-ft tall building as shown
in the following figure.
The height of the rocket is given by y(t)=150t -16. 1t2 + 50 ft.
Find the following:
a) The velocity, v(t)=
as well as the initial velocity v0 = v(0).
Solution:
v(t)=
= 150 -32.2t ft/sec
v(0) = 150 ft/sec
b) The acceleration a(t) =
Solution:
a(t) =
=
=
= -32.2 ft/sec2
c) The time required to reach the maximum height as well as the corresponding
height ymax. Use your results to sketch y(t).
Solution:
The maximum height occurs when v(t)=0
v(t) = 150-32.2t=0
32.2t = 150 t=4.66 sec
y(max) = y(4.66) -16.1(4.66)2 + 50 = 399 ft
The ball hits the ground when y(t)=0
150t -16. 1t2+50 = 0
solving, t=-0.32 or 9.64 seconds
t=-0.32 is not reasonable
6. (30 points) A current flowing through a capacitor is shown graphically
in the following figure.
If we know that i(t) =
=
, sketch the graphs of the stored charge
and the voltage, assuming C=250 μF, q(0)=0 milli coulombs and V(0)=0 volts.
Solution:
Stored charge, q(t),
= i(t) = slope of the charge plot
0
t
1 ms
=1(
)
q(t) = linear
1
t
2 ms
=0(
)
q(t) = constant
2
3
4
t
3 ms
=2(
)
t
q(t) = linear
4 ms
=1
t
q(t) = linear
5 ms
= -2
q(t) = - linear
Voltage , v(t)
i(t) =
= i(t) =
Note:
is the same as
but scaled by
Thus for each time interval, the slope of q(t) is simply scaled by for v(t).
0
t
1 ms
(1) = 4
1
t
2 ms
(0) = 0
2
t
3 ms
(2) = 8
3
t
4 ms
(1) = 4
4
t
5 ms
(-2) = -8