AP Calculus AB · Lesson 1

Writing differential equations and reading slope fields

Know Know it

A differential equation links an unknown function to its derivative, such as dydt=0.3y\frac{dy}{dt} = 0.3y. A solution is a function that makes the equation true. To check one, find its derivative and substitute both into the equation.

Rate sentences translate directly. "The rate of change of y is proportional to Q" means dydt=kQ\frac{dy}{dt} = kQ for a constant k. "Proportional to the difference between T and 20" means dTdt=k(T−20)\frac{dT}{dt} = k(T - 20). The sign of the whole right side says whether the quantity is rising or falling at that moment.

A slope field draws, at each grid point (x,y)(x, y), a short segment whose slope is the value of dydx\frac{dy}{dx} there. A solution curve runs along the segments. Segments are horizontal wherever dydx=0\frac{dy}{dx} = 0.

If dydx\frac{dy}{dx} depends only on x, every segment in a vertical column has the same slope. If it depends only on y, every segment in a horizontal row has the same slope.

Trap Columns versus rows

For dydx=x2\frac{dy}{dx} = x^2, the slope changes as you move left or right but not as you move up or down, so the segments match down each column. Students often say "the same along each row", which is the pattern for an equation in y alone, such as dydx=y2\frac{dy}{dx} = y^2.

Worked example Easy

The number of fish N in a pond grows at a rate proportional to the number of fish present. Which differential equation models this, where k is a positive constant and t is time?

  1. dNdt=kt\frac{dN}{dt} = kt
  2. N=ktN = kt
  3. dNdt=kN\frac{dN}{dt} = kNAnswer
  4. dNdt=kN\frac{dN}{dt} = \frac{k}{N}

How to solve it

  1. "The rate at which N grows" is dNdt\frac{dN}{dt}.
  2. "Proportional to N" means equal to a constant times N, so dNdt=kN\frac{dN}{dt} = kN.

Why each choice is right or wrong

  • A. Incorrect. dNdt=kt\frac{dN}{dt} = kt makes the rate of growth depend on how much time has passed, not on how many fish there are.
  • B. Incorrect. N=ktN = kt describes the number of fish itself growing in a straight line; it says nothing about a rate proportional to N.
  • C. Correct. The rate of change is dNdt\frac{dN}{dt}, and "proportional to the number present" makes it a constant k times N.
  • D. Incorrect. dNdt=kN\frac{dN}{dt} = \frac{k}{N} is inversely proportional to N, so more fish would mean slower growth.

Worked example Medium

Which statement describes the slope field for the differential equation dydx=x−y\frac{dy}{dx} = x - y?

  1. Every segment on the x-axis is horizontal.
  2. Every segment on the line y=xy = x is horizontal.Answer
  3. All segments in the same vertical column have the same slope.
  4. Every segment has a positive slope.

How to solve it

  1. A segment is horizontal where dydx=0\frac{dy}{dx} = 0.
  2. x−y=0x - y = 0 exactly when y=xy = x, so the flat segments lie on the line y=xy = x.
  3. Above that line y>xy > x, so the slope is negative; below it the slope is positive.

Why each choice is right or wrong

  • A. Incorrect. On the x-axis y=0y = 0, so the slope is x−0=xx - 0 = x, which is not zero except at the origin.
  • B. Correct. On the line y=xy = x, dydx=x−x=0\frac{dy}{dx} = x - x = 0, so each segment there is flat.
  • C. Incorrect. The slope x−yx - y changes with y, so segments in one vertical column have different slopes.
  • D. Incorrect. Below the line y=xy = x the slope x−yx - y is positive, but above that line it is negative.

Explain Explain it

Task verbs: explain, determine

To explain what a slope field shows at a point, substitute the point into dydx\frac{dy}{dx} and say the sign: positive means the solution curve is rising there, negative means falling, zero means a flat segment.

For concavity, differentiate the equation with respect to x. Treat y as a function of x, so the chain rule gives a factor of dydx\frac{dy}{dx} wherever y appears. Then substitute the formula for dydx\frac{dy}{dx} and the point. A positive second derivative means concave up; a negative one means concave down.

Worked example Hard

A solution curve of dydx=xy−1\frac{dy}{dx} = xy - 1 passes through the point (1,3)(1, 3). What is the value of d2ydx2\frac{d^2y}{dx^2} at that point?

  1. 66
  2. 22
  3. 33
  4. 55Answer

How to solve it

  1. Differentiate xy−1xy - 1 with the product rule, treating y as a function of x: d2ydx2=y+xdydx\frac{d^2y}{dx^2} = y + x\frac{dy}{dx}.
  2. At (1,3)(1, 3): dydx=(1)(3)−1=2\frac{dy}{dx} = (1)(3) - 1 = 2.
  3. So d2ydx2=3+(1)(2)=5\frac{d^2y}{dx^2} = 3 + (1)(2) = 5, and the curve is concave up there.

Why each choice is right or wrong

  • A. Incorrect. The value 6 uses dydx=xy=3\frac{dy}{dx} = xy = 3 and drops the −1-1 from the equation, giving 3+33 + 3.
  • B. Incorrect. The value 2 is the first derivative dydx\frac{dy}{dx} at the point, not the second derivative.
  • C. Incorrect. The value 3 treats y as a constant, so the derivative of xyxy becomes just y and the xdydxx\frac{dy}{dx} term is lost.
  • D. Correct. d2ydx2=y+xdydx\frac{d^2y}{dx^2} = y + x\frac{dy}{dx}. At (1,3)(1, 3), dydx=3−1=2\frac{dy}{dx} = 3 - 1 = 2, so the value is 3+1(2)=53 + 1(2) = 5.

Explain Your turn

A solution curve of dydx=y(4−y)\frac{dy}{dx} = y(4 - y) passes through the point (0,1)(0, 1). Explain whether the curve is increasing or decreasing at that point, and whether it is concave up or concave down there. Show the values you use.

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Compare with a model answer

A full answer does this

  • Uses the slope at (0,1)(0, 1), dydx=1(4−1)=3>0\frac{dy}{dx} = 1(4 - 1) = 3 > 0, to conclude the curve is increasing there
  • Shows that d2ydx2\frac{d^2y}{dx^2} is positive at (0,1)(0, 1) (it equals 6) and concludes the curve is concave up there

Model answer

At (0,1)(0, 1), dydx=1(4−1)=3\frac{dy}{dx} = 1(4 - 1) = 3. The slope is positive, so the curve is increasing there. Writing the equation as dydx=4y−y2\frac{dy}{dx} = 4y - y^2 and differentiating with respect to x gives d2ydx2=(4−2y)dydx\frac{d^2y}{dx^2} = (4 - 2y)\frac{dy}{dx}. At (0,1)(0, 1) this is (4−2)(3)=6(4 - 2)(3) = 6, which is positive, so the curve is concave up there.

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