Answer:
![\frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3D%5Cfrac%7B6y%7D%7Bx%7D%5Ctext%7B%20ft%20per%20sec%7D)
Step-by-step explanation:
Let L be the length of the ladder,
Given,
x = the distance from the base of the ladder to the wall, and t be time.
y = distance from the base of the ladder to the wall,
So, by the Pythagoras theorem,
![L^2 = y^2 + x^2](https://tex.z-dn.net/?f=L%5E2%20%3D%20y%5E2%20%2B%20x%5E2)
,
Differentiating with respect to time (t),
![\frac{dL}{dt}=\frac{d}{dt}(\sqrt{x^2 + y^2})](https://tex.z-dn.net/?f=%5Cfrac%7BdL%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%28%5Csqrt%7Bx%5E2%20%2B%20y%5E2%7D%29)
![=\frac{1}{2\sqrt{x^2 + y^2}}\frac{d}{dt}(x^2 + y^2)](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%5Csqrt%7Bx%5E2%20%2B%20y%5E2%7D%7D%5Cfrac%7Bd%7D%7Bdt%7D%28x%5E2%20%2B%20y%5E2%29)
![=\frac{1}{2\sqrt{x^2 + y^2}}(2x\frac{dx}{dt}+2y\frac{dy}{dt})](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%5Csqrt%7Bx%5E2%20%2B%20y%5E2%7D%7D%282x%5Cfrac%7Bdx%7D%7Bdt%7D%2B2y%5Cfrac%7Bdy%7D%7Bdt%7D%29)
![=\frac{1}{\sqrt{x^2 +y^2}}(x\frac{dx}{dt}+y\frac{dy}{dt})](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%5E2%20%2By%5E2%7D%7D%28x%5Cfrac%7Bdx%7D%7Bdt%7D%2By%5Cfrac%7Bdy%7D%7Bdt%7D%29)
Here,
![\frac{dy}{dt}=-6\text{ ft per sec}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3D-6%5Ctext%7B%20ft%20per%20sec%7D)
Also,
( Ladder length = constant ),
![\implies \frac{1}{\sqrt{x^2 +y^2}}(x(-6)+y\frac{dy}{dt})=0](https://tex.z-dn.net/?f=%5Cimplies%20%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%5E2%20%2By%5E2%7D%7D%28x%28-6%29%2By%5Cfrac%7Bdy%7D%7Bdt%7D%29%3D0)
![-6x + y\frac{dy}{dt}=0](https://tex.z-dn.net/?f=-6x%20%2B%20y%5Cfrac%7Bdy%7D%7Bdt%7D%3D0)
![y\frac{dy}{dt}=6x](https://tex.z-dn.net/?f=y%5Cfrac%7Bdy%7D%7Bdt%7D%3D6x)
![\implies \frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}](https://tex.z-dn.net/?f=%5Cimplies%20%5Cfrac%7Bdy%7D%7Bdt%7D%3D%5Cfrac%7B6y%7D%7Bx%7D%5Ctext%7B%20ft%20per%20sec%7D)
Which is the required notation.