A clock has a 15cm hour hand and a 20cm hour hand. Find how fast the distance is changing between the two hands when it is 3:00 p.m.
Let the distance be a,
a^2 = b^2 + c^2 -2bc (cos A)
a^2 = 15^2 + 20^2 -2(15)(20)(cos A)
a^2 = 625 - 600(cos A)
Differentiating,
2aa'= 600(sin A) x dA/dt
dA/dt = -11pi/6 rad/hr (i'm not sure if this is right)
so plugging in values,
2(25)a' = 600(-11pi/6)
50a' -1100pi
Thus, a' = -22pi cm/hr
i'm not sure if this is right and im wondering if anyone can verify with me
Let the distance be a,
a^2 = b^2 + c^2 -2bc (cos A)
a^2 = 15^2 + 20^2 -2(15)(20)(cos A)
a^2 = 625 - 600(cos A)
Differentiating,
2aa'= 600(sin A) x dA/dt
dA/dt = -11pi/6 rad/hr (i'm not sure if this is right)
so plugging in values,
2(25)a' = 600(-11pi/6)
50a' -1100pi
Thus, a' = -22pi cm/hr
i'm not sure if this is right and im wondering if anyone can verify with me