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If $ x^2 + y^2 + z^2 = 9, dx/dt = 5, $ and $ dy/dt = 4, $ find $ dz/dt $ when $ (x, y, z) = (2, 2, 1). $
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01:51
Wen Zheng
Calculus 1 / AB
Chapter 3
Differentiation Rules
Section 9
Related Rates
Derivatives
Differentiation
Missouri State University
Baylor University
Boston College
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.
44:57
In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.
04:22
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So in this problem, the first thing what we want to focus on is just looking at what we have. So we have X squared bus y squared a Z squared People's Nine and ultimately, this problem is helping us to do a better job of understanding related rates problems. So what we have here is, uh, we're gonna want to start by taking the derivative. So when we differentiate with respect to T that is on both sides, we'll end up getting two x times dx DT Class two I times d y d t class choosy. I'm dizzy E t. And we know this is equal to zero because we differentiate the nine. And with all this in mind, we end up seeing that X is equal toe one are actually X is equal to two. But we can factor out the two out of here. That makes it much easier. We know the X is equal to two. Why is equal to two z is equal to one so we can substitute those. We know that DX DT is equal to five d y d t is. You go to form and we're trying to find these e D t. So, since this we're multiplying by one here, we can just simplify that. That this is eight. This is 10. So 10 plus eight is 18. So now we can put this over here. When we subtract it, it'll be negative. 18. So what we've just done is found busy, bt, um, which essentially means we found the derivative of Z with respect to t. Um and we know that to be 18, a negative 18.
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