Question 9
The slope field for the differential equation $\dfrac{dy}{dx}=y-x^2$ is shown.
The solution curve to the differential equation that passes through the point $(3,-5)$ also passes through point
That’s it. Step through the hints below to see why. In 2021, 58.0% of Queensland students chose (A).
Not quite. Q is almost straight above $(3,-5)$, and the segments there look nearly vertical. They lean back to the left, though. At $(3,-5)$ the gradient is $-5-3^2=-14$, so as the curve climbs it moves left, away from Q. The curve through Q drops off the bottom of the grid near $x\approx3.9$. In 2021, 10.0% of Queensland students chose (B).
Not quite. Left of P the segments are still steep with negative gradient, so moving left the curve keeps climbing. At $x=-4.1$ it is up near $y\approx10.6$, far above R. R is on a different curve, which drops off the bottom of the grid near $x\approx-3.7$. In 2021, 7.0% of Queensland students chose (C).
Not quite. S is the closest point to $(3,-5)$, but the curve climbs too steeply to pass through it. At $x=1.3$ the curve through $(3,-5)$ is up near $y\approx2.3$, above S. The curve through S drops off the bottom near $x\approx2.1$. In 2021, 24.7% of Queensland students chose (D).
The blue curve is the solution through $(3,-5)$. Pick another point to draw its solution curve in black. The slider moves the orange dot along the chosen curve, with its tangent segment.
At $(3,-5)$ the gradient is $-14$, so the curve climbs steeply as you move left. It peaks near $(1.5,\,2.3)$, dips to about $0.5$ near $x=-0.7$, then climbs again through P.
The curve through P is the same curve as the one through $(3,-5)$. That is option (A).
The curve through Q drops off the bottom of the grid near $x\approx3.9$, to the right of $(3,-5)$, so it misses that point.
The curve through R drops off the bottom near $x\approx-3.7$. At $x=-4.1$ the blue curve is up near $y\approx10.6$, far above the grid.
The curve through S drops off the bottom near $x\approx2.1$. At $x=1.3$ the blue curve is up near $y\approx2.3$, above S.
To the left, the solution curves crowd together, so this one also passes very close to P. What matters is that it misses $(3,-5)$.
A solution curve follows the segments. The gradient at any point is $y-x^2$. Start at $(3,-5)$.
$$\frac{dy}{dx}=-5-3^2=-14$$
A negative gradient means the curve rises as you move left. From $(3,-5)$ it goes steeply up and to the left. Moving right, it falls off the bottom of the grid.
Keep following the segments left. The curve levels off near $(1.5,\,2.3)$, dips to about $0.5$ near $x=-0.7$, then climbs again. Near $x=-2.9$ it reaches P at about $y=4.5$. Check the gradient there:
$$\frac{dy}{dx}\approx4.55-(-2.9)^2\approx-3.86$$
That matches the segments around P, which slope down to the right.
At $(3,-5)$ the gradient is $-5-9=-14$, so the curve climbs steeply to the left. Following the segments, it levels off across the middle of the field and then climbs again, passing through P.
Q, R and S each sit on a different solution curve.
How Queensland students answered in 2021
From the QCAA Specialist Mathematics subject report for 2021. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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