y = x^0.5
dy /dx = 0.5 x^-0.5
dy/dt = (0.5 /sqrtx) dx/dt
if x = 16
dy/dt = ( 0.5 / 4 ) * 4 cm/s = 0.5
v = sqrt (dx/dt^2 + dy/dt^2)
= sqrt (4^2 + .25) = sqrt(16.25)
A particle is moving along the curve below.
y = sqroot(x)
As the particle passes through the point (16, 4), its x-coordinate increases at a rate of 4 cm/s. How fast is the distance from the particle to the origin changing at this instant? (Round your answer to three decimal places.)
2 answers
the distance z can be found using
z^2 = x^2 + y^2 = x^2 + x
2z dz/dt = (2x+1) dx/dt
so at (16,4)
2√260 dz/dt = (2*16+1)(4)
dz/dt = 132/(2√260) = 33/√65
z^2 = x^2 + y^2 = x^2 + x
2z dz/dt = (2x+1) dx/dt
so at (16,4)
2√260 dz/dt = (2*16+1)(4)
dz/dt = 132/(2√260) = 33/√65
3 answers