6.1 Notes - Slope Fields

Greg Kelly, Hanford High School, Richland, Washington

AP Calculus AB/BCAP Calculus AB/BC

First, a little review:

Consider:2 3y x

then: 2y x 2y x

2 5y x or

It doesn’t matter whether the constant was 3 or -5, since when we take the derivative the constant disappears.

However, when we try to reverse the operation:

Given: 2y x find .y

2y x C

We don’t know what the constant is, so we put “C” in the answer to remind us that there might have been a constant.

If we have some more information we can find C.

Given: and when , find the equation for .2y x y4y 1x

2y x C 24 1 C

3 C2 3y x

This is called an initial value problem. We need the initial values to find the constant.

An equation containing a derivative is called a differential equation. It becomes an initial value problem when you are given the initial condition and asked to find the original equation.

Example 1. Find the general solution to the exact differential equation.

4 25 sec dy

x xdx

First, find the antiderivative of .dy

dx

5 tan y x x C

This is the general solution to the Differential Equation (D.E.). If the general solution is continuous, a unique solution can be found if an initial condition if given. This gives a particular solution to the D.E.

Example 2. Find the particular solution to the equation.

3 sin , and 2 when 0. dy

x y xdx

3 sindy

xdx

1. Find the general solution by finding the antiderivative.

3 cos y x C 2. Substitute the initial condition.

2 3 cos 0 C 3. Solve for C.

2 3 1 C

5C

4. Substitute C into the general solution to get the particular solution.

3 cos 5 y x

Example 3. Your turn. Find the particular solution to the equation.

6 27 3 5, and 1 when 1. du

x x u xdx

6 27 3 5 du

x xdx

7 3 5 u x x x C

7 31 1 1 5 1 C 4C

7 3 5 4 u x x x

Example 4. Solve the initial value problem explicitly.

2 4

1 312, and 3 when 1.

dyy x

dx x x

If a function is discontinuous, then the initial condition only pins down the continuous piece of the curve that passes through the given point. In this case the domain must be specified.

The point (1, 3) only pins down the continuous piece over a

certain interval. This interval is (0, 8). When x = 1,

This interval needs to be taken into account as part of the

solution. Since the point (1, 3) only takes into account positive

values for x, the domain is

8.dy

dx

0, .

2 4

1 312, and 3 when 1.

dyy x

dx x x

Example 4. Solve the initial value problem explicitly.

2 43 12 dy

x xdx

3

1 112 y x C

x x

3

1 13 12 1

1 1 C 11C

3

1 112 11, Domain 0 y x x

x x

Sometimes we can’t easily find the antiderivative, so we need to use the Fundamental Theorem of Calculus.

Example 5. Solve the initial value problem using the Fundamental Theorem.

2sin , and 5 when 1. dy

x y xdx

2

1sin x

y t dt C

1 2

15 sin t dt C

5 0 C 5C

2

1sin 5 x

y t dt

Initial value problems and differential equations can be illustrated with a slope field.

Slope fields are mostly used as a learning tool and are mostly done on a computer or graphing calculator, but a recent AP test asked students to draw a simple one by hand.

General solutions always have that + C added on as part of the function and when graphing y = f(x) + C.

The + C shifts the graph vertically by whatever the value of C is and gives us a family of functions.

Draw a segment with slope of 2.

Draw a segment with slope of 0.

Draw a segment with slope of 4.

2dy

xdx

x ydy

dx

0 0 00 1 00 00 0

23

1 0 21 1 2

2 0 4

-1 0 -2

-2 0 -4

2dy

xdx

If you know an initial condition, such as (1,-2), you can sketch the curve.

By following the slope field, you get a rough picture of what the curve looks like.

In this case, it is a parabola.

Example 5. Construct a slope field for the differential equation and sketch a graph of the particular solution that passes through the point (2, 0).

dy

x ydx

x ydy

dx0 0 01 -1 0

-1 1 01 0 12 0 23 0 30 1 10 -1 -11 -2 -12 -3 -1

Since (2, 0) is above what appears to be an asymptote, we only need to sketch the slope field above the asymptote to find the particular solution.

Integrals such as are called definite integrals

because we can find a definite value for the answer.

4 2

1x dx

4 2

1x dx

43

1

1

3x C

3 31 14 1

3 3C C

64 1

3 3C C

63

3 21

The constant always cancels when finding a definite integral, so we leave it out!

=

=

=

Integrals such as are called indefinite integrals

because we can not find a definite value for the answer.

2x dx

2x dx31

3x C

When finding indefinite integrals, we always include the “plus C”.

=

A graphing calculator can be used to graph the integral of a function using the numerical integration function (fnInt).

1 fnInt sin , ,0,y x x x xor0

sin x

y t t dtThis is extremely slow and usually not worth the trouble.

[-10,10] by [-10,10]