AP Slope Fields Worksheet Key
AP Slope Fields Worksheet
4
Slope fields give us a great way to visualize a family of antiderivatives,
solutions of differential equations.
dy
Solving
dx
y
3
2
= 2 x means “Name a function whose derivative is 2x”.
1
2
2
2
Answers might include y = x , y = x + 3 , y = x − 4 , and so forth.
x
−4 −3 −2 −1
−1
In general, y = x + C .
If you sketch several of these antiderivatives on the same graph grid you
see the family of antiderivatives.
Another way to show the family of antiderivatives is to draw a slope field
2
1
2
3
4
5
−2
−3
−4
5
dy = 2 x . The small segments simply represent the slope of the functions at various points. It gives the
for
dx
shape of the possible solutions for the differential equations. Look at the slope field and visualize the family of
antiderivatives. You can also sketch the solution curve through a particular point..
Drawing a Slope Field
You are simply drawing a short “line”, given a point and a slope.
Intro. Draw a slope field for a differential equation such as dy dx = 2 .
Pick a starting point on the grid and draw a tiny line segment that passes
through the point and that has a slope of 2.
Move to another point, the slope will also be 2.
y
2
1
x
−3
−2
−1
1
2
3
4
−1
Summarize & Analyze: The slope will be 2 no matter what point we
consider since dy dx = constant .
−2
1) Draw a slope field for the differential equation dy dx = x + 1 .
2
Pick a starting point on the grid, say (0, 0), find the slope
dy dx = 0 + 1 = 1 , and draw a tiny line segment that passes through the
point and with the slope that you found.
Name other points that have the same slope. Draw the segments.
Repeat the process with another point, until the slope field is filled.
Summarize & Analyze: All of the points with the same
x-coordinate have the same slope, because the
differential equation has an x-term but no y-term.
x
–2
–1
0
1
2
y
any
any
any
any
any
1
x
−3
−2
−1
S. Stirling 2011-12
1
2
3
4
−1
dy/dx
–1
0
1
2
3
−2
2) Draw a slope field for dy dx = 2 y
Repeat the process established in part b) until the slope field is filled.
Summarize & Analyze: All of the points that have the
same y-coordinate have the same slope because
this differential equation contains a y-term but no
x-term.
y
2
y
1
x
any
any
any
any
any
y
–2
–1
0
1
2
dy/dx
–4
–2
0
2
4
x
−3
−2
−1
1
−1
−2
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2
3
4
AP Slope Fields Worksheet Key
Simple Slope Field Patterns Summary:
If all line segments on the slope field have the same slope, then the differential equation will be of the form
dy/dx = constant.
If all line segments in the vertical direction on a slope field have the same slope, then the differential equation
does not contain a y-term. The x-coordinate determines the slope.
If all segments in the horizontal direction on a slope field have the same slope, then the differential equation
does not contain an x-term. The y-coordinate determines the slope.
With differential equations that contain both an x-term and a y-term, such as dy/dx = x + y, look for points that
have the same slope.
Draw a slope field for each of the following differential equations. Each tick mark is one unit.
3) dy dx = x + y
4) dy dx = 2 x
2
y
2
y
1
1
x
x
−3
x
0
–2
1
2
–2
y
–2
0
2
2
2
−2
dy/dx
–2
–2
3
4
0
−1
1
2
3
4
−3
−2
−1
1
−1
−1
−2
−2
Same slope when x and y
interchanged. Zero when
opposites.
Segments get steeper as abs. of x
and y increase.
QI all positive; QIII all negative.
5) dy dx = y − 1
x
0
–2
2
1
–1
y
any
any
any
any
any
dy/dx
0
–4
4
2
–2
2
3
Zero when x = 0.
Same slope in vertical direction
& opposites when reflected over
the x-axis. QI & QIV all
positive; QII & QIII all negative.
6) dy dx = − y / x
2
y
2
1
y
1
x
−3
x
any
any
any
any
any
y
–2
–1
0
1
2
−2
dy/dx
–3
–2
–1
0
1
S. Stirling 2011-12
4
−1
1
2
3
x
4
−3
−2
−1
1
−1
−1
−2
−2
Zero when y = 1.
Same slope in horizontal
direction. Steeper slope as y
moves away from y = 1.
Positive y > 1 & negative y < 1.
x
2
–4
0
–1
2
y
–4
2
any
1
2
dy/dx
2
1/2
0
1
–1
2
3
4
Zero when y = 0. Undefined
when x = 0. Inverse points are
reciprocals.
Slope of 1 opposite coordinates.
Slope of –1 same coordinates.
Page 2 of 7
AP Slope Fields Worksheet Key
Match the slope fields with their differential equations.
7) dy dx = sin x
8) dy dx = x − y
9) dy dx = 2 − y
(C) Flow appears to
make the pattern of
the neg. cosine
curve. Oscillates.
(D) Zero when x = y.
Pos. x > y &
Neg. x < y .
(A) Horizontally
same. No x-term.
Zero when y = 2.
10) dy dx = x
(B) Vertically
same. No y-term.
Zero when x = 0.
AP: Match
slope fields to
equations
(solution).
(all a scale of 1)
Look for graph
within the field
“super impose”
Look for
symmetries and
say what
matched.
Match the slope fields with their differential equations.
11) dy dx = 0.5 x − 1
12) dy dx = 0.5 y
13) dy dx = − x / y
(B) Vertically
same. No y-term.
Zero when x = 2.
S. Stirling 2011-12
(C) Horizontally
same. No x-term.
Zero when y = 0.
(D) Neg. 1 when x = y
Pos. 1 when x = –y.
Zero when x = 0.
Undefined when y = 0.
14) dy dx = x + y
(A) Zero when
opposites x = –y.
Pos. x > –y.
Page 3 of 7
AP Slope Fields Worksheet Key
15) The slope field from a certain differential equation is shown
above. Which of the following could be a specific solution to that
differential equation?
2
x
(B) y = e
(A) y = x
(E) y = ln x
−x
(C) y = e
(D) y = cos x
(E)
Almost vertical toward y-axis.
Levels off as x →∞.
16) The slope field for a certain differential equation is shown. Which
of the following could be a specific solution to that differential
equation? D
(A) y = sin x
(B) y = cos x
2
(C) y = x
(D) y =
1 3
x
6
(D)
Almost vertical as x →± ∞.
Zero when x = 0. Odd function = symmetry
AP: now differential equations
= depends on x so col alike
check slopes match at a couple
look for horizontal tangents
S. Stirling 2011-12
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AP Slope Fields Worksheet Key
17. Consider the differential equation given by
dy xy
=
dx 2
(A) On the axes provided, sketch a slope field for the given differential equation.
(B) Let f be the function that satisfies the given differential equation. Write an equation for the tangent line to
the curve y = f ( x) through the point (1, 1). Then use your tangent line equation to estimate the value of f
(1.2).
(C) Find the particular solution y = g ( x) to the differential equation with the initial condition g (1) = 1 . Use
your solution to find g(1.2).
(D) Compare your estimate of f (1.2) found in part (B) to the actual value of g(1.2) found in part (C). Was your
estimate from part (B) an underestimate or an overestimate? Use your slope field to explain why.
x
2
2
–2
–2
4
4
–4
–4
y
2
1
x
−3
−2
−1
1
2
3
4
−1
y
1
–1
–1
1
1
–1
–1
1
dy/dx
1
–1
1
–1
2
–2
2
–2
Zero when x = 0 or y = 0.
Symmetric about x-axis &
y-axis.
−2
(B)
dy xy
=
dx 2
dy 1
= x dx
y 2
1
1
∫ ydy = 2 ∫ x dx
(C)
e
ln y
=e
(D) The estimate in part (B) was an under
estimate. Since the graph of g is concave up the
tangent line lies below the curve.
1 x2 +C1
4
y = ±eC1 ie
y = Ce
1 x2
4
2
1 x2
4
1 (1)2
4
,
1
e
1
g ( x) =
e
g (1.2) =
1
y
1
Initial cond.: g (1) = 1
1 = Ce
dy (1)(1) 1
1
1
1
=
= so y = ( x − 1) + 1 or f ( x) ≈ x +
dx
2
2
2
2
2
1
1
f (1.2) ≈ (1.2 ) + = .6 + .5 ≈ 1.1
2
2
e
1
=C
x
−3
−2
−1
1
2
3
4
−1
4
−2
1 x2
4
4
1
e
1
S. Stirling 2011-12
e
1 (1.2)2
4
≈ 1.116
4
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AP Slope Fields Worksheet Key
18. Consider the differential equation given by
(A)
(B)
(C)
(D)
(E)
dy x
=
dx y
On the axes provided, sketch a slope field for the given differential equation.
Sketch a solution curve that passes through the point (0, 1) on your slope field.
Find the particular solution y = f ( x) to the differential equation with the initial condition f (0) =1.
Sketch a solution curve that passes through the point (0, −1) on your slope field.
Find the particular solution y = f ( x) to the differential equation with the initial condition f (0) = −1.
2
y
1
x
−3
−2
−1
1
2
3
−1
4
x
2
2
–2
–2
3
3
–3
–3
y
1
–1
–1
1
1
–1
–1
1
dy/dx
2
–2
2
–2
3
–3
3
–3
Zero when x = 0
Undefined when y = 0.
One when x = y
Neg one when x = – y
(B) sketch:
2
−2
y
1
x
dy x
=
(C)
dx y
y dy = x dx
−3
OR Initial cond.: f (0) = 1
−2
−1
1 2 1 2
(1) = (0) + C1
2
2
1
C1 =
2
1 2 1 2 1
y = x +
So
2
2
2
2
2
y = x +1
∫ y dy = ∫ x dx
1 2 1 2
y = x + C1
2
2
2
2
y = x + C2
Initial cond.: f (0) = 1
12 = 02 + C2 , C2 = 1
1
2
−1
−2
y 2 = x 2 + 1 and y = ± x2 + 1
since y is positive, choose the
positive part.
y = x2 + 1
(D)
2
(E)
2
2
Since y = x + C2
Initial cond.: f (0) = −1
( −1)
2
= 0 + C2 , C2 = 1
2
So y = x + 1 and
since y is negative, choose the
2
2
y
1
x
−3
−2
−1
1
2
3
4
−1
−2
positive part, y = − x + 1
2
S. Stirling 2011-12
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4
AP Slope Fields Worksheet Key
2006 AB 5 (No calculator)
y
2
y
dy 1 + y
=
Consider the differential equation
, where
dx
x
x ≠ 0.
(a) On the axes provided, sketch a slope field for the
given differential equation at the eight points
indicated. (scale is 1)
2 pts: sign of
slope at each point
and relative
steepness of slope
lines in rows and
columns.
x
y
dy/dx
-2
0
-1/2
-1
0
-1
1
0
1
2
0
1/2
all
-1
0
-1
1
-2
1
1
2
11
x
x
−2
−1
1
−1
1
2
2
−1 −1
2 −2
Note x ≠ 0 !
Only draw at the dots indicated!
Sign of slopes correct and relative
steepness of the slopes of the lines in the
rows and columns.
(b) Find the particular solution y = f ( x ) to the differential equation with the initial condition
f ( −1) = 1 and state its domain.
Written as “should be a
function”, f(x).
dy 1 + y
=
dx
x
dy
dx
=
1+ y x
ln 1 + y = ln x + C1
1 + y = ±eC1 ie
ln x
y = C x −1
1 pt: separates variables
2 pts: antiderivatives
1 pt: constant of integration
1 pt: uses initial condition
1 pt: solves for y
Note: Max 3/6 if no constant of integration
Note: 0/6 if no separation of variables
Should split and pick a branch.
y = C x − 1 for negative x.
⎧ u, u > 0
u =⎨
⎩−u, u < 0
y = C ( − x ) −1
Initial condition: f (−1) = 1
so 1 = C ( − ( −1) ) − 1 ,
1 = C −1, C = 2
So y = 2 x − 1 when x < 0 ,
OR if you replaced x = − x for x < 0
y = −2 x − 1 when x < 0
both accepted MUST have domain!
1 pt: domain
S. Stirling 2011-12
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