A rectangle has one side of 8 cm. How fast is the diagonal of the rectangle changing at the instant when the other side is 6 cm and increasing at 3 cm per minute?
So the diagonal formula is $$ sqrt(w^2+l^2)$$ We could say x = 8. Y= 6. dy/dt = 3cm per minute. How would we set this up from here?
Best Answer
So unless your teacher is mandating you use x and y I would recommend sticking with D, w, and l, just so things don't get confusing.
That being said you have started the equation perfectly with
$ D = \sqrt{ (w^2 + l^2)} $
From here you should implicitly derive the equation
$ \frac{dD}{dt} = \frac{2*w*\frac{dw}{dt}+2*l*\frac{dl}{dt}}{2\sqrt{w^2 + l^2}} $
Here I will let l = 8, dl/dt = 0, w = 6, and dw/dt = 3. Now you substitute that into the implicitly derived equation
$ \frac{dD}{dt} = \frac{2*6*3 + 2*8*0}{2\sqrt{6^2 + 8^2}} $
$ \frac{dD}{dt} = \frac{36}{2\sqrt{100}} = 1.8 $
I hope that makes sense