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
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
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
To find the rate of change of the distance from the particle to the origin, we can use the Pythagorean theorem. Let's call the distance from the particle to the origin "d". We can express "d" as:
d = √(x^2 + y^2)
where x is the x-coordinate and y is the y-coordinate of the particle.
Now, let's differentiate both sides of the equation with respect to time (t):
d/dt = d/dt (√(x^2 + y^2))
To find this derivative, we can apply the chain rule. Remember that y is a function of x, so we need to differentiate it with respect to t. The chain rule states that:
d/dt (√(x^2 + y^2)) = (∂/∂x (√(x^2 + y^2))) * dx/dt + (∂/∂y (√(x^2 + y^2))) * dy/dt
Let's calculate these partial derivatives:
∂/∂x (√(x^2 + y^2)) = (1 / (2√(x^2 + y^2))) * 2x = x / √(x^2 + y^2)
∂/∂y (√(x^2 + y^2)) = (1 / (2√(x^2 + y^2))) * 2y = y / √(x^2 + y^2)
Now, let's substitute these partial derivatives back into our differentiation equation:
d/dt (√(x^2 + y^2)) = (x / √(x^2 + y^2)) * dx/dt + (y / √(x^2 + y^2)) * dy/dt
We are given that dx/dt = 5 units per second. To find dy/dt, we need to differentiate y with respect to t. Since we don't have an equation for y in terms of t, we can use the fact that x and y are related by the curve equation y = 2 √(4x + 4). Let's differentiate this equation implicitly with respect to t:
dy/dt = (∂y/∂x) * (dx/dt)
To find (∂y/∂x), we differentiate y with respect to x:
(∂y/∂x) = ∂/∂x (2 √(4x + 4)) = 2 * (1 / (2√(4x + 4))) * 4 = 2 / √(4x + 4)
We substitute this back into the differentiation equation for dy/dt:
dy/dt = (2 / √(4x + 4)) * (dx/dt)
Now, we can substitute both dx/dt and dy/dt back into the differentiation equation for d/dt:
d/dt (√(x^2 + y^2)) = (x / √(x^2 + y^2)) * 5 + (y / √(x^2 + y^2)) * (2 / √(4x + 4)) * 5
Given that the particle passes through the point (3, 8), we can substitute x = 3 and y = 8 into this equation to find the rate of change of the distance from the particle to the origin at that instant.