Question 7
The diagram shows a slope field.
The differential equation represented by the slope field is
Not quite. With $\frac{dy}{dx}=x+y$ you get a zero gradient along $y=-x$, not along $y=x$.
Not quite. $y-x$ gives you zero along the correct line $y=x$, but you have the sign reversed on each side. Hint 3 shows this at $(0.75,-0.75)$.
That’s it. Step through the hints below to see why.
Not quite. With $\frac{dy}{dx}=x$ you ignore $y$ entirely, so every segment in a vertical column would be parallel, and the horizontal segments would lie along the $y$-axis ($x=0$).
Black: the question’s slope field. Orange: the slope field of the option you are testing. Dashed blue: where that option gives a gradient of 0. Tap a segment, or use the sliders, to move the blue test ring.
The orange segments are horizontal along $y=-x$, but the diagram’s are horizontal along $y=x$. This slope field doesn’t match.
The orange segments are horizontal along $y=x$, like the diagram’s, but off that line each one slopes the opposite way. This slope field doesn’t match.
Every orange segment sits under a black one. This slope field matches the diagram.
This field ignores $y$, so every segment in a vertical column is parallel, and the gradient is 0 only along the $y$-axis ($x=0$). This slope field doesn’t match.
Find where the segments are horizontal. In the diagram they lie along the line $y=x$, for example at $(0.75,0.75)$ and $(-1.25,-1.25)$.
$$\frac{dy}{dx}=0\ \text{ along }\ y=x$$
Horizontal segments mark where $\frac{dy}{dx}=0$. Setting each option to zero gives a line you can match against the diagram, which is the fastest way to eliminate options.
Find which options are zero along $y=x$.
$$\begin{aligned}&\text{(A)}\quad x+y=0\ \Rightarrow\ y=-x\quad\text{Not }y=x\\&\text{(B)}\quad y-x=0\ \Rightarrow\ y=x\quad\text{Possible}\\&\text{(C)}\quad x-y=0\ \Rightarrow\ y=x\quad\text{Possible}\\&\text{(D)}\quad x=0\text{ is the }y\text{-axis}\quad\text{Not }y=x\end{aligned}$$
Only (B) and (C) remain.
Only those options can match the diagram.
Test the sign of the gradient at one point off the line. Take the plotted point $(0.75,-0.75)$, which is below the line $y=x$. In the diagram, the segment there (like every segment below $y=x$) slopes upwards, so the gradient there is positive.
$$\begin{aligned}&\text{(C)}\quad \frac{dy}{dx}=0.75-(-0.75)=1.5>0\\&\text{(B)}\quad \frac{dy}{dx}=-0.75-0.75=-1.5<0\end{aligned}$$
(C) matches. (B) does not match.
(B) and (C) are negatives of each other, $y-x=-(x-y)$, so one point decides between them.
The segments are horizontal along $y=x$, which leaves (B) and (C). At $(0.75,-0.75)$ the segment slopes upwards, and only $x-y$ is positive there.
$$\frac{dy}{dx}=x-y$$
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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