# Lesson 4.1 day 2

advertisement ```4.1 Antiderivatives and Indefinite Integration
Day 2
Initial Value Problems
the problem of finding a function when we know its derivative function and the value y = F(x) for one value of x
initial condition
to solve:
1.
Find the general solution, y = F(x) + C from the differential equation, 2.
dy
= f (x)
dx
Find the right value of C by using the initial condition 1
Example:
Find the curve whose slope is 3x2 if the curve is required to pass through the point (1, &shy;1).
differential equation:
initial condition:
...
curve:
2
(more)
Examples:
Slope Fields A2
4.1 Exercise 54 (page 250) [gray handout]
(a) Sketch two approximate solutions . . . (b) Use integration to find the particular solution . . .
differential equation:
dy
dx
= x2 &shy; 1
point:
(&shy;1, 3)
3
p250 #62. Solve the differential equation.
f &quot;(x) = sin x, f '(0) = 1, f (0) = 6
4
Applications
[salmon handout]
1) (p 250 &shy; Exercise #64) Population Growth The rate of growth dP/dt of a population of bacteria is proportional to the square root of t, where P is the population size and t is the time in days (0 t 10). That is
dP
dt
=k t
.
The initial size of the population is 500. After 1 day the population has grown to 600. Estimate the population after 7 days.
5
recall:
s(t) position function
v(t) = s ' (t) velocity function
a(t) = v ' (t) = s &quot; (t) acceleration function
note:
a(t) acceleration function
v(t) = a(t) dt velocity function
s(t) = v(t) dt = a(t) dt position function
6
Applications
[salmon handout]
Solving a Vertical Motion Problem A ball is thrown upward with 2) (p 248 &shy; Example 8)
an initial velocity of 64 feet per second from an initial height of 80 feet. (See Figure 4.4). a.) Find the position function giving the height s as a function of the time t.
b.) When does the ball hit the ground?
7
Assignment
p250-252 #49 - 63 odd,
#65 - 69,
#71 - 75
8
```