Video: Identifying the Slope Field Given a Differential Equation

Which of the following is the slope field of the differential equation 𝑦′ = 𝑒^(2π‘₯) + 3? [A] Slope field A [B] Slope field B [C] Slope field C [D] Slope field D [E] Slope field E

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Video Transcript

Which of the following is the slope field of the differential equation 𝑦 prime equals 𝑒 to the power of two π‘₯ plus three?

Remember, slope fields, which are sometimes also called vector fields or direction fields, are a way of graphically obtaining the solutions to first-order differential equations. Our differential equation is 𝑦 prime equals 𝑒 to the power of two π‘₯ plus three. Now, we could rewrite this if needs be in Leibniz form such that d𝑦 by dπ‘₯ equals 𝑒 to the power of two π‘₯ plus three. This part is very much a matter of personal choice. To work out which of the graphs is the slope field of the differential equation we have. We’re going to substitute values of π‘₯ into our differential equation to establish the slope at each point.

It’s really sensible to do this in some sort of logical manner. A table can really help us organize our thoughts. A sensible choice is to begin with an π‘₯ value of zero. d𝑦 by dπ‘₯ in this case is 𝑒 to the power of two time zero or 𝑒 to the power of zero plus three. Now of course, 𝑒 to the power of zero is one. So when π‘₯ is equal to zero, d𝑦 by dπ‘₯ or 𝑦 prime is equal to four. This is greater than zero. We also see that the expression for the slope is independent of 𝑦. So it will be four for all values of π‘₯. And this actually means we can disregard two of our slope fields immediately. Notice for slope fields B and E, when π‘₯ is equal to zero, the slope is negative. So we’ll disregard these.

We’ll now try π‘₯ equals one. When π‘₯ equals one, 𝑦 prime or d𝑦 by dπ‘₯ is 𝑒 squared plus three. That’s approximately equal to 10.3. Once again, 10.3 is positive. And that means we can disregard a further slope field. In this slope field, when π‘₯ is equal to one, the slope is negative. So we need to decide between slope fields A and D. We could try π‘₯ equals two. That gives us a value for 𝑦 prime of 𝑒 to the fourth power plus three, which is approximately 23.1, steeper still. Now, we can actually see that as π‘₯ increases, so does 𝑦 prime. And it will always be positive for positive values of π‘₯.

So let’s try negative values of π‘₯. What about when π‘₯ is equal to negative one? 𝑦 prime is 𝑒 to the power of negative two plus three? That’s approximately 3.1, still positive but slightly less steep than the slope when π‘₯ is equal to zero. We’ll try one more value of π‘₯. Let’s let π‘₯ be equal to negative two. 𝑦 prime is an 𝑒 to the negative four plus three, which is roughly 3.0, only slightly less steep than at π‘₯ equals negative one, but still positive. Now in fact, let’s think about part of our derivative 𝑒 to the power of two π‘₯. It’s always greater than zero.

This means 𝑒 to the power of two π‘₯ plus three will always be greater than three. No matter the value of π‘₯, our slope will always be greater than three. And this means we can eliminate one further slope field. In slope field A, the slope is either zero or something very close to zero for values of π‘₯ less than zero. We disregard this fourth slope field. Let’s just double check D satisfies our criteria. 𝑦 prime is always greater than zero, although for values of π‘₯ less than zero it doesn’t seem to change considerably, whereas for values of π‘₯ greater than zero, we can see it grows pretty quickly. The correct slope field is, therefore, this one. In this example, it’s D but on your screen, it might be in a different order.

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