y = 2√(4x+4) = 4√(x+1)
THE DISTANCE D IS FOUND USING
d^2 = x^2+y^2 = x^2+16(x+1) = x^2+16x+16
when x=3, d=√73
so,
2d dd/dt = (2x+16) dx/dt
d dd/dt = (x+8) dx/dt
so, now for the numbers:
√73 dd/dt = 11(5)
dd/dt = 55/√73
A particle is moving along the curve y= 2 \sqrt{4 x + 4}. As the particle passes through the point (3, 8), its x-coordinate increases at a rate of 5 units per second. Find the rate of change of the distance from the particle to the origin at this instant.
*I just need step by step process in how I am able to find dy/dt and dd/dt and how to solve this out. I am aware I need to go with derivative, i'm just confused how to set it up
2 answers
looks like you defined d to be the distance of the point from the origin
y = 2(4x+4)^(1/2)
y^2 = 4(4x+4) = 16x + 16
d = √(x^2+y^2)
= ( x^2 + 16x + 16)^(1/2
dd/dt = (1/2)(x^2 + 16x + 16)^(-1/2) (2x + 16) dx/dt
so when x = 3, y = 8 and dx/dt = 5 . Plug away
dd/dt = (1/2) (1/√73)(22)(5)
=appr 6.44 units per second
check my arithmetic
y = 2(4x+4)^(1/2)
y^2 = 4(4x+4) = 16x + 16
d = √(x^2+y^2)
= ( x^2 + 16x + 16)^(1/2
dd/dt = (1/2)(x^2 + 16x + 16)^(-1/2) (2x + 16) dx/dt
so when x = 3, y = 8 and dx/dt = 5 . Plug away
dd/dt = (1/2) (1/√73)(22)(5)
=appr 6.44 units per second
check my arithmetic