Question

Please help Ladder question!!

Answer 1

Answer:

Step-by-step explanation:

This is a related rates problem from calculus using implicit differentiation. The main equation is going to be Pythagorean's Theorem and then the derivative of that. Pythagorean's Theorem is

$x^2+y^2=c^2$ where c is the hypotenuse and is a constant. Therefore, the derivative of this with respect to time, and using implicit differentiation is

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$ and dividing everything by 2 to simplify a bit:

$x\frac{dx}{dt}+y\frac{dy}{dt}=0$. Upon analyzing that equation, it looks like we need values for x, y, $\frac{dx}{dt}$, and $\frac{dy}{dt}$. And here's what we were given:

$\theta=45$ and $\frac{dx}{dt}=.5$  In the greater realm of things, that's nothing at all.

BUT we can use the right triangle and the angle we were given to find both x and y. The problem we are looking to solve is to

Find $\frac{dy}{dt}$ at the instant that $\frac{dx}{dt}$ = .5.

Solving for x and y:

$tan45=\frac{x}{6}$ and

6tan45 = x ( and since this is a 45-45-90 triangle, y = x):

$6(\frac{\sqrt{2} }{2})=x=y$ so

$x=y=3\sqrt{2}$ and now we can fill in our derivative. Remember the derivative was found to be

$x\frac{dx}{dt}+y\frac{dy}{dt}=0$ so

$3\sqrt{2}(\frac{1}{2})+3\sqrt{2}\frac{dy}{dt}=0$ and

$\frac{3\sqrt{2} }{2}+3\sqrt{2} \frac{dy}{dt}=0$ and

$3\sqrt{2}\frac{dy}{dt}=-\frac{3\sqrt{2} }{2}$ and multiplying by the reciprocal of the left gives us:

$\frac{dy}{dt}=-\frac{3\sqrt{2} }{2}(\frac{1}{3\sqrt{2} })$ so

$\frac{dy}{dt}=-\frac{1}{2}\frac{m}{s}$