# Question Video: Determining the Correct Slope Field Graph of a Given Differential Equation Mathematics • Higher Education

Which of the following is the slope field of the differential equation 𝑦′ = 𝑥²? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

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### Video Transcript

Which of the following is the slope field of the differential equation 𝑦 prime is equal to 𝑥 squared? Options (A), (B), (C), (D), or (E).

We’re given five potential sketches of the slope field for the differential equation 𝑦 prime is equal to 𝑥 squared. And we need to determine which is the correct sketch. To start, we’re going to need to recall what a slope field is for a differential equation. We recall the slope field of a differential equation means at the point 𝑥, 𝑦, we plot the slope d𝑦 by d𝑥 at 𝑥, 𝑦. In other words, we’re substituting 𝑥 and 𝑦 into the equation for our slope. In this case, we’re given that 𝑦 prime is equal to 𝑥 squared. So we need to recall that 𝑦 prime is the same as d𝑦 by d𝑥. In other words, in this case, d𝑦 by d𝑥 is equal to 𝑥 squared.

So to determine the sketch of the slope field of this differential equation, we need to substitute values of 𝑥 into this differential equation. Let’s start with the origin when 𝑥 is equal to zero and 𝑦 is equal to zero. And it’s also worth pointing out here, since d𝑦 by d𝑥 is equal to 𝑥 squared, which is entirely a function in 𝑥, our inputs of 𝑦 won’t change the value of our slope. And we can just calculate this. We substitute 𝑥 is equal to zero into d𝑦 by d𝑥 is equal to 𝑥 squared. This gives us the slope of zero squared which is of course equal to zero. So our slope at the origin is equal to zero.

In fact, this would also have been true if we had chosen 𝑦 equal to one or any value of 𝑦. So, in fact, our slope should be equal to zero along the entire vertical line 𝑥 is equal to zero. So let’s plot this onto one of our sketches. Let’s plot this onto sketch (A). We know when the slope is zero, we will have horizontal lines, so we need to plot these along our entire line 𝑥 is equal to zero. And we can see that all five of our sketches have this property. So this doesn’t help us eliminate any of our options.

So we’re going to need to try more values of 𝑥 and 𝑦. Let’s try 𝑥 is equal to one, and once again we know that this will be true for any value of 𝑦. We substitute 𝑥 is equal to one into our expression for d𝑦 by d𝑥. This gives us d𝑦 by d𝑥 is equal to one squared, which of course simplifies to give us one. This time, we need to plot lines of slope one along the entire line of 𝑥 is equal to one. But now we can see something interesting. For example, in plot (A), we can see when 𝑥 is equal to one, our lines approximately have slope negative one.

The same thing is true in plot (B). We can see when 𝑥 is equal to one, all of our lines have slope negative one. So options (A) and (B) cannot be correct. These lines are supposed to have slope positive one. All three of our remaining options have positive slope lines when 𝑥 is equal to one. We might be tempted to eliminate option (C) since when 𝑥 is equal to one, it appears the slope of our lines are less than one. However, we’ll refrain from doing this because this is a sketch.

We can now try inputting more values. Let’s now try 𝑥 is equal to negative one. And once again, we know the value of 𝑦 won’t change our slope. So we input 𝑥 is equal to negative one into our slope function of 𝑥 squared, giving us negative one all squared. And, of course, negative one all squared is equal to one. So on the line 𝑥 is equal to negative one, we need to have a slope of positive one. And now we can see something interesting. For example, in sketch (C), we can see when 𝑥 is equal to negative one, all of the slopes have negative slope.

The same is also true in sketch (D). We can see all of the slopes when 𝑥 is equal to negative one are negative. But we know they’re supposed to be equal to positive one, so option (C) and option (D) cannot be correct. This means that option (E) must be correct since we’ve eliminated all of our other options. However, let’s just check that this is true. We can start by sketching the three slopes we’ve already found. Let’s start by sketching the slopes of zero when 𝑥 is equal to zero. And we can see that our sketch does indeed have this property.

We can do the same for the two other slopes we found. When 𝑥 is equal to one and 𝑥 is equal to negative one, we should have slopes equal to positive one. And we can see this is also true in our sketch. When 𝑥 is equal to one and when 𝑥 is equal to negative one, all of our slopes are approximately equal to positive one. Finally, we can think what happens when we increase our value of 𝑥 or when we decrease our value of 𝑥, for example, when 𝑥 is equal to 10 or when 𝑥 is equal to negative 10. We know in both of these instances, our value of 𝑦 will not change our derivative. If we were to evaluate the derivative in both of these cases, we would get 10 squared and negative 10 all squared, respectively.

And the thing here to notice is as we increase our value of 𝑥 from one, our derivative increases. Similarly, as we decrease our value of 𝑥 from negative one, our slope also increases. And we can then ask the question, what would this mean for the sketch of our slope field? As we increase our value of 𝑥 from one, our slopes need to get larger and larger. We can see this happens in our sketch. Similarly, as we decrease our value of 𝑥 from negative one, our slopes also need to get larger and larger. And we can see this also happens in our sketch.

And this is our final answer. In this question, we were given five possible sketches of the slope field of the differential equation 𝑦 prime is equal to 𝑥 squared. And we were able to show that only option (E) was a possible sketch of the slope field of this differential equation.