Problem

A particle is moving along a hyperbola $ xy = 8. …

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Problem 11 Easy Difficulty

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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Video Transcript

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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Problem

A particle is moving along a hyperbola $ xy = 8. …

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Caleb Elmore

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Catherine Ross

Caleb Elmore

Joseph Lentino

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Caleb Elmore

Joseph Lentino

James Stewart

Calculus: Early Transcendentals