Answer:
<em>The x-coordinate is changing at 10 cm/s</em>
Step-by-step explanation:
<u>Rate of Change</u>
Suppose two variables x and y are related by a given function y=f(x). If they both change with respect to a third variable (time, for instance), the rate of change of them is computed as the derivative using the chain rule:
![\displaystyle \frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdt%7D%3D%5Cfrac%7Bdy%7D%7Bdx%7D%5Ccdot%20%5Cfrac%7Bdx%7D%7Bdt%7D)
We have
![y=\sqrt{8+x^3}](https://tex.z-dn.net/?f=y%3D%5Csqrt%7B8%2Bx%5E3%7D)
Or, equivalently
![y^2=8+x^3](https://tex.z-dn.net/?f=y%5E2%3D8%2Bx%5E3)
We need to know the rate of change of x respect to t. We'll use implicit differentiation:
![\displaystyle 2y\cdot \frac{dy}{dt}=3x^2\cdot \frac{dx}{dt}](https://tex.z-dn.net/?f=%5Cdisplaystyle%202y%5Ccdot%20%5Cfrac%7Bdy%7D%7Bdt%7D%3D3x%5E2%5Ccdot%20%5Cfrac%7Bdx%7D%7Bdt%7D)
Solving for dx/dt
![\displaystyle \frac{dx}{dt}=\frac{2y\cdot \frac{dy}{dt}}{3x^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdx%7D%7Bdt%7D%3D%5Cfrac%7B2y%5Ccdot%20%5Cfrac%7Bdy%7D%7Bdt%7D%7D%7B3x%5E2%7D)
Plugging in the values x=1, y=3, dy/dt=5
![\displaystyle \frac{dx}{dt}=\frac{2(3)\cdot 5}{3\cdot 1^2}=10](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdx%7D%7Bdt%7D%3D%5Cfrac%7B2%283%29%5Ccdot%205%7D%7B3%5Ccdot%201%5E2%7D%3D10)
The x-coordinate is changing at 10 cm/s
X - is negative, and Y positive , so this is quadrant II (second quadrant)
Answer:
You had to solve the equation to it's fullest (so work out the problem on paper) Apperently that is her version of explaining something and I lost 100 braincells reading that! XD
Step-by-step explanation:
Hope this helps! :)