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
[tex]x^2+y^2=c^2[/tex] where c is the hypotenuse and is a constant. Therefore, the derivative of this with respect to time, and using implicit differentiation is
[tex]2x\frac{dx}{dt}+2y\frac{dy}{dt}=0[/tex] and dividing everything by 2 to simplify a bit:
[tex]x\frac{dx}{dt}+y\frac{dy}{dt}=0[/tex]. Upon analyzing that equation, it looks like we need values for x, y, [tex]\frac{dx}{dt}[/tex], and [tex]\frac{dy}{dt}[/tex]. And here's what we were given:
[tex]\theta=45[/tex] and [tex]\frac{dx}{dt}=.5[/tex] 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 [tex]\frac{dy}{dt}[/tex] at the instant that [tex]\frac{dx}{dt}[/tex] = .5.
Solving for x and y:
[tex]tan45=\frac{x}{6}[/tex] and
6tan45 = x ( and since this is a 45-45-90 triangle, y = x):
[tex]6(\frac{\sqrt{2} }{2})=x=y[/tex] so
[tex]x=y=3\sqrt{2}[/tex] and now we can fill in our derivative. Remember the derivative was found to be
[tex]x\frac{dx}{dt}+y\frac{dy}{dt}=0[/tex] so
[tex]3\sqrt{2}(\frac{1}{2})+3\sqrt{2}\frac{dy}{dt}=0[/tex] and
[tex]\frac{3\sqrt{2} }{2}+3\sqrt{2} \frac{dy}{dt}=0[/tex] and
[tex]3\sqrt{2}\frac{dy}{dt}=-\frac{3\sqrt{2} }{2}[/tex] and multiplying by the reciprocal of the left gives us:
[tex]\frac{dy}{dt}=-\frac{3\sqrt{2} }{2}(\frac{1}{3\sqrt{2} })[/tex] so
[tex]\frac{dy}{dt}=-\frac{1}{2}\frac{m}{s}[/tex]
