Video Transcript
Which of the following is the slope field of the differential equation d𝑦 by d𝑥 is equal to 𝑥 cubed times the sin of 𝑥? Option (A), (B), (C), (D), or (E).
In this question, we’re given five possible sketches of the slope field of the differential equation d𝑦 by d𝑥 is 𝑥 cubed times the sin of 𝑥. We need to determine which one is the correct sketch. To do this, we’re first going to need to recall what the slope field of a differential equation is. And we recall to sketch the slope field of a differential equation at the point 𝑥, 𝑦, we just need to plot the slope. And we’re given an equation for the slope: the slope d𝑦 by d𝑥 is equal to 𝑥 cubed times the sin of 𝑥.
So to find the values of d𝑦 by d𝑥, we’re going to need to substitute values of 𝑥 into this function. We can do this however we want. One sensible method is to use a table. But before we start substituting values of 𝑥, there’s a couple of things we need to notice. First, we see 𝑦 is not included in our differential equation. This tells us in the sketch of our slope field of this differential equation, the value of 𝑦 will not affect our slope. In other words, if we move vertically, the slope of all of our lines should be the same.
And now there’s one more thing we need to discuss before we start substituting in values of 𝑥. We see that our differential equation includes the sin of 𝑥. Recall this is a differential equation. So because we include trigonometric functions, these will all be in radians, since whenever we’re doing calculus involving trigonometric functions, we’re always using radians. Now, we’re ready to start substituting in values of 𝑥.
Let’s start with 𝑥 is equal to zero. Substituting this into our differential equation, we get zero cubed times the sin of zero. And of course, we can evaluate this to get zero. So when 𝑥 is equal to zero, our slope d𝑦 by d𝑥 is equal to zero. And remember, this will be true for any value of 𝑦. We need to recall a line of slope zero will be a horizontal line. The rate of change of 𝑦 is equal to zero with respect to 𝑥. To see what this means to our slope field sketch, let’s plot this on the sketch (A).
We’ve shown when 𝑥 is equal to zero, our slopes are all equal to zero, and this is true for any value of 𝑦. So when the entire line 𝑥 is equal to zero, we need to have horizontal slope lines. And we can see this is true in sketch (A). In fact, this is true for all five of our sketches, so we can’t eliminate any options at the moment. Let’s try substituting 𝑥 is equal to one. Substituting 𝑥 is equal to one into our differential equation, we get d𝑦 by d𝑥 is equal to one cubed times the sin of one. And if we evaluate this, we get approximately 0.84. Remember, we’re evaluating this in radians. We see this is a positive number approximately equal to one. So, our slope lines will look a lot like the line 𝑦 is equal to 𝑥.
We now need to see which of our five sketches have this property. Let’s start with sketch (A). We’ve shown when 𝑥 is equal to one and for all values of 𝑦, our slope lines are supposed to be about 0.84. In sketch (A), we can see that we have positive slope lines; however, we’re not sure on the exact value of their slope. If we were to estimate this, we would say it’s probably less than 0.84. However, because this is a sketch, we won’t discount this possibility yet. Let’s instead move on to sketch (B). We can see when 𝑥 is equal to one, all of the lines have slope approximately equal to one.
However, if we move on to sketch (C), we can see when 𝑥 is equal to one, all of our slopes have negative value. So, (C) cannot be the slope of the sketch field of this differential equation. If it were, we wouldn’t need to have positive slope lines on 𝑥 is equal to one. Checking sketch (D), we can see that all of the lines on 𝑥 is equal to one have positive slope. So, option (D) could still be correct. And finally, when we check option (E), we can see when 𝑥 is equal to one, all of the lines have negative slope. So, option (E) cannot be correct. Since when 𝑥 is equal to one, all of our slope lines should be approximately equal to 0.84.
We still need to eliminate more options, so we need to try new values of 𝑥. Let’s try 𝑥 is equal to negative one. Substituting 𝑥 is equal to negative one into our differential equation, we get d𝑦 by d𝑥 is equal to negative one all cubed multiplied by the sin of negative one. And if we evaluate this, we get approximately 0.84. So, in actual fact, when 𝑥 is equal to negative one, we have the same situation. We should have lines of slope approximately 0.84. And this time when we check sketch (A), we can see when 𝑥 is equal to negative one, all of our lines have negative slope. And we know when 𝑥 is equal to negative one, the slope is supposed to be positive. This tells us that (A) cannot be our sketch.
If we then check sketch (B), we can see when 𝑥 is equal to negative one, all of our slopes do indeed have positive slope. So, sketch (B) still satisfies this condition. However, if we check our last option of option (D), we can see when 𝑥 is equal to negative one, our slopes are all negative. So, sketch (D) can’t possibly be our sketch. When 𝑥 is equal to negative one, the slopes are all supposed to be positive. This means we’ve eliminated all of our options. And, in fact, option (B) must be the correct sketch.
Therefore, given five possible sketches of the slope field of the differential equation d𝑦 by d𝑥 is equal to 𝑥 cubed times the sin of 𝑥, we were able to determine that option (B) must be the correct sketch. We did this by substituting values of 𝑥 into our differential equation to determine information about the slope field.
