Video: AP Calculus AB Exam 1 β€’ Section I β€’ Part B β€’ Question 78

What is the shortest distance from the origin to the graph of 𝑦 = 9/π‘₯?

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Video Transcript

What is the shortest distance from the origin to the graph of 𝑦 equals nine over π‘₯?

Now, it’s simply not enough to answer this question by sketching out the graph of 𝑦 equals nine over π‘₯ and trying to observe where the shortest distance from the origin may lie. Instead, we’ll begin by recalling the distance formula. And this tells us that for two points in the Cartesian plane β€” π‘₯ one, 𝑦 one and π‘₯ two, 𝑦 two β€” the distance between them is given by the positive square root of π‘₯ two minus π‘₯ one squared plus 𝑦 two minus 𝑦 one squared.

We can find an equation for the distance between any point on our curve and the origin by substituting zero, zero for one of the coordinates and π‘₯, 𝑦 for the other. So the distance between any coordinate on our curve and the origin is π‘₯ minus zero squared plus 𝑦 minus zero squared. And that simplifies to the square root of π‘₯ squared plus 𝑦 squared.

Remember though, the equation for our curve is 𝑦 equals nine over π‘₯. So we can substitute 𝑦 for nine over π‘₯. And we see that the distance between any point on our curve given in terms of π‘₯ is the square root of π‘₯ squared plus nine over π‘₯ all squared. We’ll simplify this a little further. And remember, we’re trying to find the shortest distance. In other words, the minimum distance from the origin to a point on the curve. What we’re going to do then is find the critical points of our function 𝑑.

Remember, we can find where the critical points of a function lie by finding the derivative of that function and setting it equal to zero. We can minimise our function for the distance then by differentiating 𝑑 with respect to π‘₯ and setting that equal to zero. So how do we differentiate 𝑑 with respect to π‘₯? Well, we rewrite our function slightly. We can say that 𝑑 is equal to π‘₯ squared plus 81π‘₯ to the negative two all to the power of one-half.

To find the derivative of 𝑑 with respect to π‘₯, we’re going to use the general power rule. This is a special case of the chain rule. And it says that if 𝑒 is some function of π‘₯, then we can differentiate 𝑒 to some constant 𝑛 with respect to π‘₯. And we get 𝑛 times 𝑒 to the power of 𝑛 minus one times d𝑒 by dπ‘₯. In our case, our 𝑒 is π‘₯ squared plus 81π‘₯ to the negative two. And d𝑒 by dπ‘₯ is equal to two π‘₯ minus 162π‘₯ to the negative three.

Substituting all of this into the formula for the general power rule, and we see that d𝑑 by dπ‘₯ is equal to a half times π‘₯ squared plus 81π‘₯ to the negative two all to the power of negative one-half multiplied by two π‘₯ minus 162π‘₯ to the negative three. We can simplify this a little. And we see that the derivative of 𝑑 with respect to π‘₯ is as shown.

Remember, we’re trying to minimise our function 𝑑. In other words, we’re trying to find the location where any critical points occur. So we’re going to set our derivative equal to zero. Now, it should be quite clear that for this fraction to be equal to zero, its numerator must be equal to zero. So two π‘₯ minus 162π‘₯ to the negative three is equal to zero. Let’s solve this for π‘₯.

We add 162π‘₯ to the negative three to both sides of our equation. We multiply both sides by π‘₯ cubed, then divide through by two. We then find the fourth root, remembering to take both the positive and the negative fourth root. And we see the critical points occur when π‘₯ is equal to positive and negative three. Now, in fact, if we look back to the function we created for the distance in terms of π‘₯, we see that substituting π‘₯ is equal to positive three and π‘₯ is equal to negative three will give us the same result. But how do we know whether we found a minimum or a maximum?

We’re going to perform the first derivative test about the point π‘₯ equals three, knowing that we’ll get the same result for π‘₯ equals negative three. We substitute π‘₯ is equal to two into the equation for d𝑑 by 𝑑π‘₯ and π‘₯ is equal to four into the equation for d𝑑 by 𝑑π‘₯. And analysing the nature of the gradient at these points will help us identify the nature of our critical point. If we substitute π‘₯ equals two, we get negative 1.649. And when we substitute π‘₯ equals four, and we get 0.595, and so on. Negative 1.649 is less than zero whereas 0.595 is greater than zero. And this tells us that the critical point at π‘₯ equals three and indeed π‘₯ equals negative three is a minimum.

We need to substitute then π‘₯ is equal to three into our formula for the distance. 𝑑 is equal to the square root of three squared plus 81 over three squared, which is three root two. And so, the shortest distance from the origin to the graph of 𝑦 equals nine over π‘₯ occurs when π‘₯ is equal to three and negative three. And this distance is three root two units.

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