Question

how would i find the rate of change of the diagonals in a rectangle? when i tried it i got D= sqrt(l^2+w^2)
then i took the dd/dt and got (.5(l^2+w^2)^-.5)(2l(dl/dt)+2w(dw/dt))

Answer 1

\[D= \sqrt{(l^2+w^2)}\]
\[D'= \frac{(l^2+w^2)'}{2\sqrt{(l^2+w^2)}}\]
\[D'= \frac{(l^2)'+(w^2)'}{2\sqrt{(l^2+w^2)}}\]
\[D'= \frac{2l+2w}{2\sqrt{(l^2+w^2)}}\]
\[D'= \frac{2(l+w)}{2\sqrt{(l^2+w^2)}}\]
\[D'= \frac{l+w}{2\sqrt{(l^2+w^2)}}\]
maybe

Answer 2

.... i forgot to delete the 2 in the denom. brains said delete and fingers said ... nah keep it

Answer 3

i dont really understand the first step

Answer 4

the first step, and i left a few things out as well just ... is simply deriving the sqrt function itself

Answer 5

and as a result; the derivate of its innards pops out

Answer 6

oh ok ty

Answer 7

let me type it up better, since i really just flubbed it, from here

\[D'= \frac{(l^2)'+(w^2)'}{2\sqrt{(l^2+w^2)}}\]
\[D'= \frac{2l\ l'+2w\ w'}{2\sqrt{(l^2+w^2)}}\]
\[D'= \frac{l\ l'+w\ w'}{\sqrt{(l^2+w^2)}}\]
is a better rendition, the 's being the derivatives or rates of change of the length and width