Video: AP Calculus AB Exam 1 β€’ Section II β€’ Part B β€’ Question 1

Consider the differential equation 𝑦 = ((1/π‘₯) (d𝑦/dπ‘₯)) βˆ’ 3. i. Draw a slope field at the four indicated points in the given figure. ii. Let 𝑦 = 𝑓(π‘₯) be a particular solution of the differential equation. Write an equation of the tangent to the graph of 𝑓 at the point (0, 2). iii. Find the approximate value of 𝑓(0.02) using the equation of the tangent line in part (ii). iv. Find 𝑦 = 𝑓(π‘₯) with 𝑓(0) = 2.

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Video Transcript

Consider the differential equation 𝑦 is equal to one over π‘₯ d𝑦 by dπ‘₯ minus three. Part i) Draw a slope field at the four indicated points in the given figure. Part ii) Let 𝑦 is equal to 𝑓 of π‘₯ be a particular solution of the differential equation. Write an equation of the tangent to the graph of 𝑓 at the point zero, two. Part iii) Find the approximate value of 𝑓 of 0.02 using the equation of the tangent line in part two. Part iv) Find 𝑦 is equal to 𝑓 of π‘₯ with 𝑓 of zero is equal to two.

For the first part of this question, we’re required to draw a slope field at the four points in the figure. These points are negative one, one; negative one, two; zero, two; and one, one. In order to draw the slope field for each of these points, we need to find the value of d𝑦 by dπ‘₯ at these points. Let’s start by writing our differential equation in terms of d𝑦 by dπ‘₯. We obtain that d𝑦 by dπ‘₯ is equal to π‘₯ multiplied by 𝑦 plus three. We start by substituting in π‘₯ equals negative one and 𝑦 equals one. We obtain that d𝑦 by dπ‘₯ is equal to negative four. We can write this value into a table.

Next, we can use π‘₯ is equal to negative one, 𝑦 is equal to two. We obtain that d𝑦 by dπ‘₯ is equal to negative five. Next, we use π‘₯ is equal to zero, 𝑦 is equal to two, giving d𝑦 by dπ‘₯ is equal to zero. Finally, we use π‘₯ is equal to one, 𝑦 is equal to one, giving us the result that d𝑦 by dπ‘₯ is equal to four. Now, we have found the values of d𝑦 by dπ‘₯ at these four points. In order to draw the slope field at these four points, we simply draw a line with the gradient of d𝑦 by dπ‘₯ at the points.

The slope field at negative one, one will have a gradient of negative four. At negative one, two, it will have a gradient of negative five. At zero, two, its gradient will be zero. And at one, one, its gradient will be four. So this completes the first part of the question.

In part two, we’re told that 𝑓 is a particular solution to the differential equation. And we’re required to find the equation of the tangent to the graph at the point zero, two. Looking at our slope field, we can see that at the point zero, two, d𝑦 by dπ‘₯ is equal to zero. Now, our tangent will be a straight line. So it will be of the form 𝑦 is equal to π‘Žπ‘₯ plus 𝑏, where π‘Ž is the gradient of the tangent and 𝑏 is the 𝑦-axis intercept of the tangent. We have just shown that at the point zero, two, the gradient d𝑦 by dπ‘₯ is equal to zero. Therefore, we know that π‘Ž will be zero in our equation. So our tangent will be of the form 𝑦 is equal to 𝑏.

Now, we’re finding the tangent of the graph at the point zero, two. Zero, two is on the 𝑦-axis. Therefore, zero, two will be the 𝑦-axis intercept of our tangent. At this point, the value of 𝑦 is two. Therefore, we can conclude that the equation of our tangent is 𝑦 is equal to two.

In part three, we need to find the approximate value of 𝑓 of 0.02 using the equation of the tangent line in part two. Now, our tangent in part two was tangential at the point zero, two. Here, the π‘₯-value is zero. In part three, we’re asked to approximate the value of 𝑓 of 0.02. Here, the π‘₯-value is 0.02. Since 0.02 is very close to zero, this means that we’re able to use the tangent from part two in order to estimate the value of 𝑓 of 0.02.

In order to do this, we simply find the 𝑦-value on our tangent when π‘₯ is equal to 0.02. Now, the equation of our tangent is 𝑦 is equal to two. And it’s not depending on π‘₯. This means that 𝑦 hold the value of two regardless of the π‘₯-value. Therefore, we can use this tangent to approximate the value of 𝑓 of 0.02 as two.

For part four of the question, we need to find 𝑦 is equal to 𝑓 of π‘₯ with 𝑓 of nought equal to two. In this part of the question, we need to find a particular solution. Therefore, we’re going to need to solve our differential equation. Now, our differential equation is of the form d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯ multiplied by β„Ž of 𝑦, where in our case 𝑔 of π‘₯ is equal to π‘₯ and β„Ž of 𝑦 is equal to 𝑦 plus three. What this means is we can use the separation of variables method in order to solve this differential equation.

We can write our differential equation as d𝑦 over 𝑦 plus three is equal to π‘₯ dπ‘₯. And now, we can integrate. We have that the integral of one over 𝑦 plus three with respect to 𝑦 is equal to the integral of π‘₯ with respect to π‘₯. We’re able to integrate π‘₯ with respect to π‘₯ quite easily. We simply increase the power of π‘₯ by one and divide by the new power. This gives us one-half π‘₯ squared. Next, we need to integrate one over 𝑦 plus three with respect to 𝑦.

Now, we know that the integral of one over 𝑦 with respect to 𝑦 is equal to the natural logarithm of the absolute value of 𝑦. Now, we could use a 𝑒 substitution here in order to show that our integral of one over 𝑦 plus three with respect to 𝑦 is equal to the natural logarithm of the absolute value of 𝑦 plus three. However, this is fairly straightforward. And so, we can just use this result here. We obtain that the natural logarithm of the absolute value of 𝑦 plus three is equal to one-half π‘₯ squared.

Now, you may be wondering where the constants of integration for our indefinite integrals have gone. And that’s because we haven’t added them in yet. We add a constant to each side of the equation: plus π‘Ž on the left and plus 𝑏 on the right. Now, we can move this constant both to the right-hand side. And we obtain a constant on the right of π‘Ž minus 𝑏. However, since π‘Ž minus 𝑏 is also a constant, we can relabel this as 𝑐. Now that we have an equation for 𝑦 in terms of π‘₯, let’s make 𝑦 the subject of this equation.

We can write both sides of the equation as powers of 𝑒. Now that the natural logarithm in the left-hand side of the equation is the inverse function of 𝑒, therefore these two functions cancel out with one another. Therefore, we’re left with the absolute value of 𝑦 plus three is equal to 𝑒 to the power of a half π‘₯ squared plus 𝑐. Since we know that 𝑒 is positive and it is simply being raised to a power, we know that the right-hand side of our equation is always positive. Therefore, we can lose the absolute value signs here, since the right-hand side of our equation is always going to be greater than zero.

Next, we can split up the right-hand side of our equation. We write it as 𝑒 to the power of one-half π‘₯ squared times 𝑒 to the power of 𝑐. Now, 𝑒 to the power of 𝑐 is simply another constant since both 𝑒 and 𝑐 are constants. Therefore, we can rewrite this as another constant 𝑑. Now, we simply subtract three from both sides of the equation. And we’re left with 𝑦 is equal to 𝑑𝑒 to the power of one-half π‘₯ squared. Now, we have nearly found the particular solution here. All that remains is to find the value of our constant 𝑑.

We’ll be using information given in the question. And that is that 𝑓 of nought is equal to two. 𝑓 of nought is equal to two tells us that at π‘₯ equals zero, 𝑦 will be equal to two. Therefore, we substitute π‘₯ equal to zero and 𝑦 equal to two into our equation. We obtain that two is equal to 𝑑 multiplied by 𝑒 to the power of zero. 𝑒 to the power of zero is simply one. Therefore, we find that 𝑑 is equal to two. From here, we reach our solution which is that 𝑦 is equal two 𝑒 to the power of one-half π‘₯ squared. We’ve now found solutions to all four parts of this question.

Part one, we found the slope field for the four points in the figure. For part two, we found that 𝑦 is equal to two. For part three, we found that 𝑓 of 0.02 is approximately equal to two. And for part four, we found that 𝑦 is equal to two 𝑒 to the power of one-half π‘₯ squared.

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