Calculus BC ยท Unit 7
Read a segment as a slope, then match the pattern of the field to a differential equation.
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A segment shows the slope at one point
At \((x,y)\), draw a short segment whose slope is \(dy/dx\) evaluated there.
Columns and rows reveal the variables
If the slope depends only on \(x\), every segment in a vertical line has the same slope.
If the slope depends only on \(y\), every segment in a horizontal line has the same slope.
Horizontal segments are equilibrium clues
The segment is horizontal where \(dy/dx=0\). A whole horizontal line of zero slopes is a constant solution.
The field is the family
Many solution curves can follow the same field. An initial point selects one curve.
Evaluate the slope at a point, or name where it is zero.
The right side does not use \(y\).
At \((2,-1)\), the slope is 2.
At \((2,3)\), the slope is 2.
The two segments match because they share an x-coordinate.
Answer: both slopes are 2
\(\dfrac{dy}{dx}=-2\)
Answer: \(-2\)
\(3-1\)
\(2\)
Answer: \(2\)
\(x=0\)
The segments on the y-axis are horizontal.
Answer: on the y-axis
A horizontal row is a fixed \(y\) and many values of \(x\).
Equal slopes along that row mean \(x\) is not affecting the slope.
Answer: A. \(y\) only
The slope is 2 at every point.
Every segment has the same steepness and the same upward direction.
Answer: every segment has slope 2
Match a picture to an equation, or read an equilibrium line.
Slope field A
Slope field B
Field A has matching segments up and down each vertical line. The slope depends only on \(x\).
On the y-axis of field A, the segments are horizontal, which matches \(dy/dx=x\).
Field B has matching segments across each horizontal line. The slope depends only on \(y\).
Above the x-axis, field B rises. Below it, field B falls. That matches \(dy/dx=y\).
Answer: A is \(dy/dx=x\), B is \(dy/dx=y\)
Above: \(y>0\), so the slopes are positive.
On the x-axis: \(y=0\), so the segments are horizontal.
Below: \(y<0\), so the slopes are negative.
Answer: positive, horizontal, negative
\((0,0)\): \(0\)
\((1,0)\): \(1\)
\((0,1)\): \(1\)
\((1,-1)\): \(0\)
Answer: 0, 1, 1, and 0
Horizontal means the right side is 0.
\(x-y=0\) on the line \(y=x\).
The other equations are not zero at every point of that line.
Answer: A. \(dy/dx=x-y\)
If \(y>2\), the slope is positive, so a solution moves up, away from 2.
If \(y<2\), the slope is negative, so a solution moves down, away from 2.
The equilibrium is unstable.
Answer: unstable
At \(x=2\), \(y=2+2=4\).
The slope is \(dy/dx=x=2\).
\(y-4=2(x-2)\)
Answer: \(y-4=2(x-2)\)
Use the field to describe the solution, not just one segment.
At every height, the slope is determined, but the starting height is not.
\(y=Ce^{x}\) threads the same field for every \(C\).
An initial condition is what selects one curve.
Answer: the field shows the whole family
The solution is \(y=3e^{-x}\).
The exponent is negative, so \(y\) decreases toward 0.
It stays positive.
Answer: decreases toward 0
Zero slope when \(y=2\).
If \(y<2\), the slope \(2-y\) is positive, so the solution increases toward 2.
The constant solution is stable from below.
Answer: \(y=2\), and yes, they move toward it
No change when \(x\) changes means the right side uses \(y\) only.
Positive above the x-axis means the sign matches \(y\).
\(dy/dx=-y\) would fall above the x-axis.
Answer: A. \(dy/dx=y\)
\(y=\dfrac{x^2}{4}+C\)
\(y(0)=0\) gives \(C=0\).
In the column \(x=2\), every segment has slope \(\dfrac{1}{2}\cdot 2=1\).
Answer: \(y=\dfrac{x^2}{4}\), column slope 1
\(dy/dx=y-3\) and \(dy/dx=(y-3)^2\) are both zero when \(y=3\).
Off that line, \((y-3)^2\) is never negative, while \(y-3\) is negative below the line.
The equilibrium line is not the whole field.
Answer: zero on that line does not fix the other slopes
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