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(a) Determine the equation of the tangent line at any point $(h, k)$ which lies on the curve $x^{2 / 3}+y^{2 / 3}=a^{2 / 3}(a>0) .$ (b) Let the tangent line to this curve intersect the $x$ and $y$ axes at the points $\left(x_{1}, 0\right)$ and $\left(0, y_{1}\right)$ respectively. Show that the length of the line segment joining these two points is a constant.

(a) $y=-\left(\frac{k}{h}\right)^{1 / 3}(x-h)+k$(b) $x_{1}=a^{2 / 3} h^{1 / 3}, y_{1}=a^{2 / 3} k^{1 / 3}$ distance $=a$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 8

Implicit Differentiation

Derivatives

Missouri State University

Oregon State University

University of Michigan - Ann Arbor

Idaho State University

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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a. Find an equation of the…

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02:10

Equations of tangent lines…

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0:00

(a) Find the slope of the …

10:46

04:10

02:28

08:24

For each implicitly define…

01:43

(a) Find an equation for t…

So this problem We've been given an equation and we want to find the tangent line at a generic point HK now because I've been given a point. Probably the easiest way to find the equation of the tangent line is to use the point slope form. Why minus why one equals m times X minus X one So x one y one There's my point HK that we've been given. The only thing I'm missing is the slope. So if we can find the slope or the derivative of my equation, we can use the point slope form. Now for this equation, we're going to use the implicit differentiation method to find the derivative. Don't forget with implicit differentiation that why is a function of X. So even though I haven't explicitly said, why equals something, we know that why is a function of X So when I take the derivative of a term with why not only am I taking the derivative of why as just a letter that I'm taking the derivative of But I have to tack on a dy dx with the chain rule because I'm taking the derivative of why. But then I have to take the derivative of why, as a function of X, So it's going to have to have that dy DX put in there. So let's look at our equation. When I take this derivative, bring down the exponents and subtract one, bring down the exponents and subtract one. But there's my wife, so I have to put on Dy DX and A is a constant, so the derivative of any constant is just zero. Now let's do a little bit of algebra to make this look a little nicer. I'm going to divide both sides by two thirds because that will get rid of my coefficients. And then I want to solve for dy DX. So I'm going to subtract X to the negative one third from both sides and divide by why to the negative one third. And I'm just going to rewrite this a little bit. I don't really want to have negative exponents. So to get rid of the negative exponents, we're going to flip numerator to denominator denominator to numerator. So that's going to give me a negative. Why is now on top? X is now on bottom and I'm racing everything to the one third power. So there's my derivative. That's the slope at any point on my, uh, on my curve. But I want the derivative at my point HK because that's what I'm going to put into my equation. So let's evaluate dy DX at the point. H k. That means my wife is okay. X is an h. So there's my slope. So if I go to put that into my equation, I'm going to have why minus y one, which is K equals my slope. Well, that's what we just found times X minus X one, which is h So that is the equation for my tangent line at a given point HK. If I know h and K, I can plug those numbers in, and then I could have this in a form that we would normally see. I put this in maybe slope intercept form. But I'm just going to leave it like this for now because of all the variables, this is probably the simplest way to write this. Okay, now that I have this, we want to examine it at two specific points. Specifically, if you imagine a line drawn on the autograph on on the grid. It's going to have an X intercept and a Y intercept. So we're gonna say the X intercept is going to happen at the point x 10 So x one is where it crosses the X axis, and it's going to have a Y intercept at zero y one. And we want to examine the distance between these two points. We'll think about the distance formula. The distance formula says I'm going to take the square root of the difference of my exes while the difference between X one and zero is just X one. So we take that difference and square it, plus the difference in my wise squared, while the difference between zero and why one. It's gonna be why one squared. And it doesn't matter if it's positive or negative. When I square it, it's going to be a positive number. So this is my distance. So if I could figure out what X one and why one were, I could put them into my distance formula and find the distance. So that's going to be our goal. We're going to try to find what x one and y one R. So I probably have to. I'm gonna have to scroll down in just a minute. But let's start with R X intercept first. I know that that intercept lies on the line that we just found in the first part. So let's plug it in. Okay. Why is going to be zero? Okay, so I have zero minus K. Well, I have negative K over age to the one third. That doesn't change. That slope is the same on the whole line. But now X is X one minus h. And now we're just gonna do a little bit of algebra to try to simplify this as much as possible. Okay, so let's and I come down here a little bit. I am going to bring, uh, this fraction over the other side, so I'm going to multiply by. It's reciprocal. So that's going to give me the negative k on the left times My reciprocal negative. Sorry. I've got the negative in the wrong spot here. I want to just make sure I'm clear the negative was outside this. So I have negative h over K to the one third. There's my reciprocal equals. X one minus h. Okay. My negatives on the left hand side. Cancel My numerator is now going to be K two, the two thirds, because it's going to cancel with the K to the one third in the denominator times H to the one third and that equals X up one minus h Add H to both sides and I get that. Now Let me factor out what they have in common. There's an H two, the one third in common. If I factor that out, I have a choice of the two thirds plus k two. The two thirds Okay, but take a look. What's in those parentheses? H two. The two thirds plus K two. The two thirds. That is the equation that we have for our line. I've got my ex and my age in place of X and y so that equals a to the two thirds so I can make that substitution. This is a church to the one third times a two the two thirds. So that's the value of X sub one. Let's do the same thing for why sub one okay, again, I apologize. This does take a lot of space. We're going to the same thing for why someone I'm going to scroll in a minute. But I want you to see that equation for the tangent line. So I have Why? I'm sorry. I'm plugging a number in here. So my why is now why someone So y sub one minus k equals the opposite of K over H two, the one third times. Well, ex Well, in this case, X is now zero. So I'm just multiplying this by negative h. Okay, so let me come up a little, So I've got some room. No, again, we're going to algebra to simplify this. My negatives on the right hand side are going to cancel. So those are those are both cancelled. I have y sub one minus k. When I multiply that that out, I now have K to the one third times h two, the two thirds when I When I cancel out that age in the one third in the denominator, I'm going to add K to both sides. And just like before, I'm going to factor out what they have in common. Both terms have a k to the one third, which leaves me with an age to the two thirds plus K two the two thirds. Well, just like last time. This is our original function. So we know that that equals a to the two thirds. So I now have a value for X up one. And why someone? Okay, now, remember, why did we do this? We did that to plug into the distance formula. So let me re copy my distance formula. It's the square root of X sub one squared plus y sub one squared. So what is X up one? It is H two, the one third times a two. The two thirds. And we're squaring that. What's my value for? Why? Okay, to the one third times a two. The two thirds and I'm squaring that. Okay, now, again, this is buddy, mostly algebra. When I square this, I have h two the two thirds times a to the four thirds K two, the two thirds a to the four thirds. But we obviously have something in common, so I'm going to factor out that a to the four thirds, what I have left is H two, the two thirds plus K two. The two thirds. Well, once again, what is this? That is our original equation. so that equals a to the two thirds. So when I put that together, I have eight of the four thirds times eight of the two thirds that is a to the six thirds or a squared. So the distance between these two, uh, intercepts is going to be a It's a constant, and we made no assumptions about where those intercepts were. We just know that no matter where they are, the distance between them is going to be the constant a.

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