### Video Transcript

Which of the following is the slope
field of the differential equation π¦ prime equals π 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 π¦
prime equals π to the power of two π₯ plus three. Now, we could rewrite this if needs
be in Leibniz form such that dπ¦ by dπ₯ equals π 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. Weβre going to substitute values of
π₯ into our differential equation to establish the slope at each point.

Itβs 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. dπ¦ by dπ₯ in this case is π to the power of two time zero or
π to the power of zero plus three. Now of course, π to the power of
zero is one. So when π₯ is equal to zero, dπ¦ by
dπ₯ or π¦ prime is equal to four. This is greater than zero. We also see that the expression for
the slope is independent of π¦. 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 weβll disregard these.

Weβll now try π₯ equals one. When π₯ equals one, π¦ prime or dπ¦
by dπ₯ is π squared plus three. Thatβs 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 π¦ prime
of π to the fourth power plus three, which is approximately 23.1, steeper
still. Now, we can actually see that as π₯
increases, so does π¦ prime. And it will always be positive for
positive values of π₯.

So letβs try negative values of
π₯. What about when π₯ is equal to
negative one? π¦ prime is π to the power of
negative two plus three? Thatβs approximately 3.1, still
positive but slightly less steep than the slope when π₯ is equal to zero. Weβll try one more value of π₯. Letβs let π₯ be equal to negative
two. π¦ prime is an π 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, letβs think about part
of our derivative π to the power of two π₯. Itβs always greater than zero.

This means π 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. Letβs just double check D satisfies
our criteria. π¦ prime is always greater than
zero, although for values of π₯ less than zero it doesnβt 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, itβs D but on your
screen, it might be in a different order.