Question

The hypotenuse of a right triangle is growing at a constant rate of a cm per second and one leg is decreasing at a constant rate of b cm per second. How fast is the acute angle between the hypotenuse and the other leg changing at the instant when both legs are 1 cm?

Answer 1

we know that dr/dt = a; dy/dt = b; and at the moment both legs are equal we have x=1; y=1; and r=sqrt(2). not all that is pertinent tho :)

Answer 2

make that: dy/dt= -b

the angle involved can be the tan(a) = y/x:

tan(a) = y/x

Dt(tan(a)) = Dt(y/x)

Answer 3

da/dt (sec^2(a)) = [ x(dy/dt) -(dx/dt)y ] / x^2

Answer 4

x(dy/dt) -(dx/dt)y
da/dt = ---------------
(xsec(a))^2

Answer 5

or should we do that with sin(a)? sin(a) = y/r... lets try that out...

r(dy/dt) - y(dr/dt)
da/dt = ----------------
x^2 cos(a)

Answer 6

x^2 should be r^2 .... my brain locked :)

Answer 7

yes thanks i think it should be sin! since we are given the opposite and the hyp

Answer 8

sqrt(2)(-b) - (1)(a)
da/dt = ---------------
sqrt(2)

Answer 9

i got 2 "a"s in that that are supposed to be different variables.... :)

Answer 10

d(theta)/dt = ......

Answer 11

x^2 + y^2 = r^2

(dx/dt)2x + (dy/dt)2y = (dr/dt)2r ; divide everything by 2

(dx/dt)x + (dy/dy)y = (dr/dt)r


(dr/dt)r - (dy/dt)y
(dx/dt) = ---------------
x

Answer 12

dx/dt = (a)(sqrt(2)) - (-b)(1)
----------------
1

Hows that look?

Answer 13

if we had the rates for "a" and "b" we'd be sittin' pretty :)

Answer 14

thanks! thats how i approached it too i just had some trouble with the simplification

Answer 15

dy/dy is sposed ta be dy/dt.... fatfingers...tiny little keyboard :)

Answer 16

I need to double check this... for x=1' y=1; r=sqrt(2)

r(dy/dt) - y(dr/dt)
da/dt = ----------------
x^2 cos(a)

sqrt(2)(-b) - (a)
-------------
(sqrt(2)/2)

that should be the answer there....

Answer 17

1^2 does not equal 2 :)

Answer 18

2(sqrt(2))(-b)/(sqrt(2)) - 2a/sqrt(2)

-2b - 2a/sqrt(2)

-2b - 2a(sqrt(2))/2

-2b - a(sqrt(2)) = d(theta)/dt

Check my math on that.... but that should be the simplified version of d(theta)

Answer 19

i beg of you ...PLEASE check my math :) if I get any more "idiot" braincells marching around in myhead, im gona scream :)