Given the equation below, find d z / d t dz/dt when d x / d t = 3 dx/dt=3 , d y / d t = 2 dy/dt=2 , x = 2 x=2 , y = 4 , y=4, and z = 1 z=1 . x 2 + y 2 − 3 z 2 = 125 x^2+y^2-3z^2=125

In recent videos, we got introduced to this idea of implicit differentiation. And recall that when doing these types of problems, we were able to take the derivative of one variable with respect to another variable using the chain rule. Now this idea is gonna be very important to understand when solving another problem type you'll see in this course called related rates. Now related rates are problems that students often find very tricky when they first learn them, but don't sweat it. Because in this video and over the course of the next few videos, I'm going to be walking you through these related rates problems you'll see in the course, including some of the more complicated ones that you'll see. And hopefully, you're gonna find this to be very straightforward because a lot of this is just related to implicit differentiation, which we've already talked about. So let's get right into things. Now whenever you have a situation where 2 or more variables are related to each other, chances are you're dealing with a related rates problem. And the first thing you wanna do with any related rates problem is you wanna see how the variables are changing with time by taking time derivatives to see how one variable affects another. Now to understand this, well, let's just get into an example here where we are given this equation, y is equal to x cubed. Now what we're asked to do here is find the rate of change, d y over d t. And notice what's also given to us in this problem is the rate of change for x, d x over d t, and we're given x at one certain instance, which is equal to 4. Now this might seem like a lot of information, and we're not really sure where to start. But like I mentioned before, a good step for any related rates problem is to take the time derivative on both sides of the equation. So what I'm going to do is take the time derivative of this entire equation up here. Now if I do this, we're literally just taking this derivative of every term. So we'll have d of dt for y, which is the derivative of y, then we'll have the time derivative of x cubed. Now the way that we can find these derivatives is by recalling what we did with implicit differentiation. And remember, when we took the derivative of one variable with respect to another, we got this kind of rate. And we know that we can use the chain rule when doing this as well, so that's what we're gonna do over here. So I know the derivative of y with respect to t is just gonna give me a d y over d t on the left side. We learned that from implicit differentiation. And I know that for taking the derivative of x cubed, I can use the power rule. Doing that is going to give me 3x squared. Now you might think that we can just finish right about here, but this is actually not the final result after taking the derivative because notice we're taking a derivative of x with respect to t. In the past when we've done this, we took the derivative of x with respect to x, which would give us a d x over d x, which naturally we don't even have to write. But since we're taking the derivative of x with respect to t, we need to use the chain rule and multiply this by d x over d t. So this right here would truly be the answer for our derivative, and that's how you can differentiate both sides with respect to time, t being the time variable here. So this would be step 1 right here, which is taking the time derivative on both sides and using implicit differentiation. Now step 2 is gonna be to isolate the target rate of change. I can see clearly in this problem the target rate of change is gonna be this dy over dt. But notice something, dy over dt is already isolated on one side of the equal sign. So we don't even need to worry about step 2. We already have this rate isolated. So really, we can just move to step 3, which tells us to plug in the known values or rates and solve. So to do this, well, let's just see what we're given in the problem. Now I can see for this equation here, we have 3 and then we have x squared. X we're told is 4 at this instance, so that's gonna be 4 squared. That's gonna be multiplied by DX over DT, which I can see right here is 2. Now I trust at this point you can do this math in your head, 3 times 4 squared times 2, and all of this should come out to 96. So this right here is going to be the rate d y over d t, and that would be the solution to this problem. So this is how you can solve these types of related rates problems. And something that I'm going to mention here is this was actually a bit more of a simple related rates problem. Because notice in this example, we were given the rate d x over d t. We were also given the equation that we need to start with, and we were asked to find the missing rate here after being directly given the values. That's not always going to happen in related rates, and that's why students oftentimes find this to be a bit trickier. But make sure to check out the next couple of videos because I'm going to walk you through some of those more complicated examples, and hopefully, you're going to see that this process is the same. See you there.

4. Applications of Derivatives

Related Rates

Calculus

4. Applications of Derivatives

Related Rates

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Multiple Choice

Given the equation below, find $d z / d t$ when $d x / d t = 3$ , $d y / d t = 2$ , $x equals 2$ , $y equals 4 comma$ and $z equals 1$ .
$x^{2} + y^{2} - 3 z^{2} = 125$

$\frac{14}{3}$

$\frac{16}{3}$

$- \frac{14}{3}$

$- \frac{16}{3}$

Verified step by step guidance

Start by identifying the given equation: $ x^2 + y^2 - 3z^2 = 125 $. This is an implicit function involving variables x, y, and z.

To find $ \frac{dz}{dt} $, we need to differentiate the given equation with respect to time t. Use implicit differentiation, applying the chain rule to each term.

Differentiate $ x^2 $ with respect to t: $ \frac{d}{dt}(x^2) = 2x \frac{dx}{dt} $. Given $ \frac{dx}{dt} = 3 $, substitute this value after differentiating.

Differentiate $ y^2 $ with respect to t: $ \frac{d}{dt}(y^2) = 2y \frac{dy}{dt} $. Given $ \frac{dy}{dt} = 2 $, substitute this value after differentiating.

Differentiate $ -3z^2 $ with respect to t: $ \frac{d}{dt}(-3z^2) = -6z \frac{dz}{dt} $. Now, substitute the known values $ x = 2 $, $ y = 4 $, and $ z = 1 $ into the differentiated equation and solve for $ \frac{dz}{dt} $.