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A point moves along the circle $x^{2}+y^{2}=5$ in such a way that the distance from the point (4,6) is increasing at a rate of 7 units per second. Find $d x / d t$ at (1,2).

-35 units/sec

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

Missouri State University

Baylor University

University of Michigan - Ann Arbor

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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Circle A point is moving …

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A point is moving along th…

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A point $P$ is moving alon…

02:12

It's always a good idea to start a related rates problem with a picture. So let's draw a picture of what we're doing here. We have a circle. Our circle is X squared, plus y squared equals five. So we're going out to the square root of five all the way around this circle. We're also told we have a 0.46 Okay, what are we looking at? We're told that we have a point on our circle, so I'm just put a point here. The point is changing, so I'm going to represent it with some variables. We'll just call it X y those air its coordinates. And I'm looking at the distance between these two points. Let's call that D It will be changing as the point X y moves around the circle. In fact, we've been told that D is changing at a rate of seven units per second. It's increasing since it's increasing its a positive seven for my rate of change. Now, what's the question I'm trying to answer here? I want to find DX DT. So that's my unknown at the 0.12 So let's take a look at what we have I have a distance and I have two points. So what equation could I use to put all these together? Well, the distance formula is an excellent place to start. Remember, the distance formula is the square root of the difference in my exes, plus the difference in my wise. So there's my equation. So before we take the derivative, let's make sure that everything is set. I have three variables d x and Y I know d d d t I want to find dx DT But that leaves me with a third variable. Why, I don't know. D y d t I'm not trying to find d y d t so if I take the derivative as is I'm gonna end up with a d Y d t. And I'm gonna have a problem because I don't know what that is. So I need to get rid of my why we know how X squared and y squared, um, relate to each other so we can rewrite this using that equation. Okay, so I can write this as well. Why squared equals five minus X squared. And if I want to solve this just for why I could take the square root. Okay, So that we can substitute in D equals the square root of X minus four squared plus the square root of five minus X squared, minus six squared. Okay, this is really messy. So what? I'm going to dio I'm gonna Before I take the derivative, I'm going to square both sides so that side gets squared and this square root is gone. That just makes us a little bit nicer. To take the derivative of You don't have to do this step. You can do it just the way it was. I find this just a little bit easier. So let's take our derivative two d d d d t, and now the right hand side. First we'll do the exes. That's too X minus four to the first times. The derivative of what's in the parentheses. DX DT. Okay, now bring down the to times what's inside those parentheses. Now we take the derivative of what is in those parentheses. Well, I have a radical in here, so that puts a two and the radical in the denominator times the derivative of what's under the radical that becomes negative. Two x d x D t. Okay. And just to make this a little bit nicer, I can get rid of those twos there. And if you see, every single term has a two, so I can also do that, So that gets just a little simpler. Okay, Now let's plug in everything we know now. We don't have a value for D. We can get it those. I'm gonna just leave a gap there for a moment. We'll come back to that D D d t iss seven. Okay, at our specific point, X is one so that parentheses becomes negative. Three dx DT is what we're looking for. Okay, X is one. So that's five minus 14 Square root of forest two to minus six is negative. Four over here. My numerator. I have negative one again over that radical five minus one. That's going to give me a two d x d t. Okay, So what is the value of D? We do have to figure that out before we could go on. Well, I confined d for a particular case so I can use this equation that we used to relate everything together for our specific case. X is one. So that's one minus four is negative. Three at squared is nine. Why is two to minus six is negative. Four square to 16. That gives me a D of five. Okay, Right. Left hand side is just 35 over here. I have negative three DX DT plus two DX DT. That just puts a negative over there multiplying everything by negative one DX DT is negative 35.

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