- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 2
Slope fields and Euler’s method practice question 9
Each graph below shows the slope field of , with the origin marked and one curve drawn on it. Which curve is the solution through ?
A solution curve follows the segments: at every point on it, its slope equals the slope of the field there.
At the slope is . Just to the right of the origin, at , it is ; just to the left, at , it is .
Theory UBC Math 100 §12.2Key results
Worked solution
Strategy
Check each curve against the field: it must pass through (0, 0), and its slope must equal x + y all along it.
Working
The solution is flat at the origin. That rules out both straight lines: y = −x has slope −1 there, and y = −x − 1 misses the origin.
The field points up just right of the origin and down just left of it, so the solution falls into the origin and climbs out: a lowest point at (0, 0), not a highest one.
A check: the bowl-shaped curve satisfies the equation at every x, and y(0) = 0.
Answer
The answer is The curve with its lowest point at the origin, rising on both sides..
Check
Differentiating the equation gives y″ = 1 + y′, which is 1 at the origin, so the solution is concave up there: a bowl, not a cap.
Where this goes wrong
Choosing the line y = −x because the segments on it are flat. Flat segments mark where solutions have zero slope as they cross; the line itself has slope −1, so it cannot be a solution.
Leans on
The derivative as the slope of a tangent line, from The definition of the derivative
Try a similar question
Exactly one of these straight lines is itself a solution curve of . Which one?Saved in this browser only.