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If $Q=x / y$ and $R=x y,$ find $d x / d t$ and $d y / d t$ if $d Q / d t=-3$ and $d R / d t=9,$ when $x=6$ and $y=3$.

$$d x / d t=-3, d y / d t=3$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

Missouri State University

University of Michigan - Ann Arbor

University of Nottingham

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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00:47

$3 x-y=9$$2 x+y=6$

for this problem. We've been given two equations. Q equals the quotient of X and y and R equals the product of X and Y. And the goal for this problem is to find both d x, d, t and D Y d t given some certain values. D Q, D, T, d, r, d, t and X and y. So before we start to solve, will be point out a couple of things. First, you'll notice that all of the derivatives have a d. T. So it's DX dy t d y d T. That means that all of the's, Q x, y and R are all functions of tea. So when we go to take the derivative, we need to use the chain rule because it's not just a derivative of X. It's the derivative of X, which is a function of T, so it's a derivative of X times dx DT. The same thing for y and Q and R. The other thing is important to note is when do we substitute were given a lot of values that we can plug into here? When do I put them in? Well, you always want to take the derivative. First we want to take the derivative of the function as it is. Okay, whatever equation were given, we take the derivative Onley after we have the derivative. Do we substitute in values to find the value at our specific point in time? Okay, so derivative first, then we substitute. So let's see what we have. Let's start with our Q functions on the left. I have D Q D, T and the right. I have a quotients. We need to use the quotient rule that is the denominator times the derivative of the numerator. Remember, we have to use the chain rule DX DT minus the numerator times the derivative of the denominator which is d y d t all over. Why squared? So there's my derivative. Let's substitute in to see what we have. De que t t is negative. Three. Why is three and again we don't have either dx DT or D Y D T. But I can plug in everything else and why Squared is going to be nine. So let me just simplify. This is much as I can. I have got negative. 27. He goes three d x d. T minus six d y d t and I can simplify that down just a little bit more and divide everything by three and I get negative. Nine equals DX DT minus two d Y d t. That's a Sfar as I could go here because I'm left with two unknowns. Three x d t and d Y d t. So let's look at our second equation. This is a product. So when I take this derivative, I'm gonna need the product rule The first times the derivative of the second plus the second times the derivative of the first again, let's substitute in. D R D t is nine excess six. Why is three again? Everything is divisible by three. So I'm going to say three equals two d y d t plus dx DT Okay, so those are my two equations Now, if you think back to algebra two equations, two unknowns, we can solve these either substitution or addition. Let's add these because they both have a negative thio. They both have a two d Y d t. One is negative and one is positive. So when I add these two negative nine plus three, that's negative. Six. The terms with the Y d d Cancel and I'm left with two dx DT oops. Which means that my value for DX DT has to equal negative three. We could take that plug that back into one of our equations. Let's just plug it into this one, since it's right here. So that's three equals two D Y d t minus three. There's my DX DT value. Add three to both sides. Divide by two and I get a value for D Y d t of three.

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