# Question Video: Finding the Length of a Parametric Equation Curve Mathematics • Higher Education

Consider the given slope field graph. Which of the following differential equations is represented on the graph? [A] π¦β² = (π₯ + 2)/(π₯ β 3) [B] π¦β² = (2 β π₯)/(π₯ β 3) [C] π¦β² = (π₯ β 2)/(π₯ + 3) [D] π¦β² = (π₯ + 2)/(3 β π₯) [E] π¦β² = (π₯ β 2)/(π₯ β 3)

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### Video Transcript

Consider the given slope field graph. Which of the following differential equations is represented on the graph? A) π¦ prime is π₯ plus two over π₯ minus three. B) π¦ prime is two minus π₯ over π₯ minus three. C) π¦ prime is π₯ minus two over π₯ plus three. D) π¦ prime is π₯ plus two over three minus π₯. Or E) π¦ prime is equal to π₯ minus two over π₯ minus three.

We can see that each of the options has a positive or negative two in the numerator and a positive or negative three in the denominator. And if we look at our graph, the behavior of the slope at π₯ is negative two and π₯ is positive three is quite distinctive. At π₯ is negative two, the slope field line segments are all horizontal. This means that the slope is equal to zero. That is, π¦ prime is equal to zero at π₯ is negative two.

So letβs try π₯ is negative two in each of our possible equations and see if we get a match. At π₯ is negative two in our equation A, π¦ prime is equal to negative two plus two over negative two plus three. Thatβs equal to zero over negative five, which is equal to zero. So equation A at the point π₯ is negative two does actually match our slope field. If we look at equation B at π₯ is negative two, we have π¦ prime is equal to two minus negative two over two minus three. Thatβs equal to four over negative five, which is not equal to zero. So equation B does not match our slope field for π₯ is negative two. And we can eliminate equation B.

Now letβs try equation C. For equation C, we have π¦ prime is equal to negative two minus two over negative two plus three. Thatβs equal to negative four over one, which is not equal to zero. So we can eliminate equation C. At π₯ is negative two in equation D, we have π¦ prime is negative two plus two over three minus negative two. Thatβs equal to zero over five, which is equal to zero. Equation D does match the slope field as π₯ is negative two. So equation D remains a contender. In equation E, we have π¦ prime is equal to negative two minus two over negative two minus three. That gives us negative four over negative five, which is four over five. This is not equal to zero. So it does not match the slope field. And we can eliminate equation E.

Weβre left now with equations A and D as possible options. Letβs now try π₯ is equal to positive three in each of these equations. At π₯ is equal to positive three, both equations have a denominator of zero. This means that both equations are undefined at π₯ is equal to three and have a vertical slope at that point. This corresponds to the slope on the slope field. So weβll have to look elsewhere, for a possible solution. Letβs choose another value, say π₯ is equal to zero. At π₯ is equal to zero, each of the line segments in the slope field has a negative slope. So letβs see what the sign of the slope is for each of our two remaining differential equations at π₯ is equal to zero.

In equation A, π¦ prime is equal to zero plus two over zero minus three, which is equal to two over negative three and is less than zero. This corresponds to the slope at π₯ is equal to zero in our slope field. So equation A is still a contender. In equation D, π¦ prime is equal to zero plus two over three minus zero. Thatβs equal to two over three, which is positive. And this does not correspond to the direction of the slope at π₯ is equal to zero in our slope field. So we can eliminate equation D. Only equation A remains. So the slope field graph represents the differential equation π¦ prime is π₯ plus two over π₯ minus three.

In this example, we were given a slope field. And we had to fit the appropriate differential equation to the graph.