Respuesta :
Answer:
[tex]\frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}[/tex]
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,
[tex]L^2 = y^2 + x^2[/tex]
[tex]\implies L = \sqrt{y^2 + x^2}[/tex],
Differentiating with respect to time (t),
[tex]\frac{dL}{dt}=\frac{d}{dt}(\sqrt{x^2 + y^2})[/tex]
[tex]=\frac{1}{2\sqrt{x^2 + y^2}}\frac{d}{dt}(x^2 + y^2)[/tex]
[tex]=\frac{1}{2\sqrt{x^2 + y^2}}(2x\frac{dx}{dt}+2y\frac{dy}{dt})[/tex]
[tex]=\frac{1}{\sqrt{x^2 +y^2}}(x\frac{dx}{dt}+y\frac{dy}{dt})[/tex]
Here,
[tex]\frac{dy}{dt}=-6\text{ ft per sec}[/tex]
Also, [tex]\frac{dL}{dt} = 0[/tex] ( Ladder length = constant ),
[tex]\implies \frac{1}{\sqrt{x^2 +y^2}}(x(-6)+y\frac{dy}{dt})=0[/tex]
[tex]-6x + y\frac{dy}{dt}=0[/tex]
[tex]y\frac{dy}{dt}=6x[/tex]
[tex]\implies \frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}[/tex]
Which is the required notation.
