Download 15-2 Derivatives and Antiderivatives

Advanced Mathematical Concepts
Chapter 15
Lesson 15-2
Example 1
a. Find an expression for the slope of the tangent line to the graph of y = x2 - 5x - 36 at any point.
dy
That is, compute .
dx
b. Find the slopes of the tangent lines when x= 0 and x = 3.
a. Find and simplify
f(x + h) - f(x)
, where f(x) = x2 - 5x - 36.
h
First, find f(x + h).
f(x + h) = (x + h)2 - 5(x + h) - 36
= x2 + 2xh + h2 - 5x - 5h - 36
Now find
Replace x with x + h in f(x).
f(x + h) - f(x)
.
h
f(x + h) - f(x) x2 + 2xh + h2 - 5x - 5h - 36 - (x2 - 5x - 36)
=
h
h
2xh + h2 - 5h
=
Simplify.
h
h(2x + h - 5)
=
Factor.
h
= 2x + h - 5
Divide by h.
dy
Now find the limit of 2x + h - 5 as h approaches 0 to compute dx.
dy
dx = f(x)
f(x + h) - f(x)
h
= lim (2x + h - 5)
= lim
h0
h0
= 2x + 0 - 5
= 2x - 5
dy
So dx = 2x - 5.
dy
b. At x = 0, dx = 2(0) - 5 or -5. The slope of the
tangent line at x = 0 is -5.
dy
At x = 3, dx = 2(3) - 5 or 1. The slope of the
tangent line at x = 3 is 1.
Advanced Mathematical Concepts
Example 2
Find the derivative of each function.
a. f(x) = x8
f(x) = 8x8 - 1
= 8x7
Power Rule
b. f(x) = x4 - 3x + 1
f(x) = x4 - 3x + 1
= x4 - 3x1 + 1
f(x) = 4x4 - 1 - 3  1x1 - 1 + 0
= 4x3 - 3x0
= 4x3 - 3
Rewrite x as a power.
Use all four rules.
x0 = 1
c. f(x) = 4x5 - 2x4 + 3x3 - 5x2 - 7x + 2
f(x) = 4  5x4 - 2  4x3 + 3  3x2 - 5  2x1 - 7  1x0 + 0
= 20x4 - 8x3 + 9x2 - 10x - 7
d. f(x) = x4 + x(4x2 - 3x)
f(x) = x4 + x(4x2 - 3x)
= x4 + 4x3 - 3x2
f(x) = 4x3 + 4  3x2 - 3  2x
= 4x3 + 12x2 - 6x
e. f(x) = (x + 4)(x2 - 3)
f(x) = (x + 4)(x2 - 3)
= x3 + 4x2 - 3x - 12
f(x) = 3x2 + 4  2x - 3 - 0
= 3x2 + 8x - 3
Multiply to write the function as a polynomial.
Chapter 15
Advanced Mathematical Concepts
Chapter 15
Example 3
ROCKETRY Refer to the application at the beginning of Lesson 15-2 in your book. Suppose the
rocket reaches its highest point 8.4 seconds after its fuel was exhausted, and it hits the ground 18.2
seconds after the fuel ran out.
a. How fast was the rocket moving at the instant its fuel ran out?
b. What was the maximum height of the rocket?
a. We have to find the value of v0. This value cannot be found directly from the height function H(t)
because H0 is still unknown. Instead we use the velocity function v(t) and what we can deduce about
the velocity of the rocket at its highest point.
H(t) = H0 + v0t - 16t2
v(t) = H(t)
= 0 + v0  1 - 16  2t
= v0 - 32t
The velocity of the rocket is the derivative of the height.
H0 and v0 are constants; t is the variable.
When the rocket was at its highest point, it was neither rising nor falling, so its velocity
was 0. Substituting v(t) = 0 and t = 8.4 into the equation v(t) = v0 - 32t yields 0 = v0 - 32(8.4), or v0 =
268.8.
The velocity of the rocket was 268.8 ft/s when the fuel ran out.
b. We can now write the equation for the height of the rocket as H(t) = H0 + 268.8t - 16t2. When the
rocket hit the ground, its height H(t) was 0, so we substitute H(t) = 0 and t = 18.2 into the height
equation.
H(t)
0
16(18.2)2 - 268.8(18.2)
H0
= H0 + 268.8t - 16t2
= H0 + 268.8(18.2) - 16(18.2)2
= H0
= 407.68
H(t) = 0, t = 18.2
Solve for H0.
The height of the rocket can now be written as H(t) = 407.68 + 268.8t - 16t2. To find the maximum
height of the rocket, which occurred at t = 8.4, compute H(8.4).
H(t) = 407.68 + 268.8t - 16t2
H(8.4) = 407.68 + 268.8(8.4) - 16(8.4)2
= 1536.64
Replace t with 8.4.
The maximum height of the rocket was about 1537 feet.
Advanced Mathematical Concepts
Chapter 15
Example 4
Find the antiderivative of the function f(x) = 4x3.
We are looking for a function whose derivative is 4x3. You may recall from previous examples that the
function x4 fits that description. The derivative of x4 is 4x4 - 1, or 4x3.
However, x4 is not the only function that works. The function G(x) = x4 + 2 is another, since the derivative
is G(x) = 4x3 + 0 or 4x3. Another answer is H (x) = x4 - 1, and still another is
J(x) = x4 + 10. Adding any constant, positive or negative, to x4 does not change the fact that the derivative
is 4x3.
All of the possible answers can be summarized by the expression x4 + C, where C is any constant. So for
the function f(x) = 4x3, we say the antiderivative is F(x) =x4 + C.
Example 5
Find the antiderivative of each function.
a. f(x) = 2x5
1
F(x) = 2  5 + 1x5 + 1 + C
Constant Multiple of a Power Rule
2
1
= 6x6 + C or 3x6 + C
b. f(x) = 3x4 - 2x3 + 1
f(x) = 3x4 - 2x3 + 1
= 3x4 - 2x3 + 1x0 Rewrite the function so that each term has a power of x.
1
1
1
F(x) = 3  x5 + C1 - 2  x4 + C2 + 1  x1 + C3 Constant Multiple of a Power Rule and
5
4
1
a Sum and Difference Rules
3 5 1 4
= x - x +x+C
Let C = C1 - C2 +C3.
5
2
c. f(x) = x3(x4 + x + 2)
f(x) = x3(x4 + x + 2)
= x7 + x4 + 2x3
Multiply to write the function as a polynomial.
1 8
1 5
1
F(x) = 8x + C1 + 5x + C2 + 2  4x4 + C3
Use all three antiderivative rules.
1
1
1
= 8x8 + 5x5 + 2x4 + C
Let C = C1 + C2 +C3.
Advanced Mathematical Concepts
Chapter 15
Example 6
PHYSICS The acceleration of a particle is given by a(t) = 4t - 9, where t is the time in seconds. The
velocity function of the particle is the antiderivative of the acceleration function.
a. Given that the particle’s velocity at 3 seconds is 120, find an equation for the velocity of the
particle.
b. Use your equation from part a to predict the particle’s velocity at 5 seconds.
a. v(t) is the antiderivative of a(t).
a(t) = 4t - 9
1
v(t) = 4  2t2 - 9t + C
= 2t2 - 9t + C
Antiderivative rules
To find C, substitute 3 for t and 120 for v(t).
120 = 2(3)2 - 9(3) + C
120 = 18 - 27 + C
C = 129
Solve for C.
Substituting this value of C into our formula for v(t) gives v(t) = 2t2 - 9t + 129. Of all the
antiderivatives of a(t), this is the only one that gives the proper velocity for the time equal to 3
seconds.
b. Substitute 5 for t.
v(t) = 2t2 - 9t + 129
v(5) = 2(5)2 - 9(5) + 129
= 134
According to the model, the velocity of the particle at 5 seconds should be about 134.