A particle is moving along the curve y = 3sqrt5x+1. As the particle passes through the point (3,12), its x-coordinate increases at a rate of 4 units per second. Find the rate of change of the distance from the particle to the origin at this instant.
I've tried several times to calculate the answers but I still can't find it. Could anyone help me with this question?
y = 3sqrt(5x+1)
dy/dt=7.5/sqrt(5x+1) * dx/dt
so dx/dt=5 at x=3, find dy/dt from last line above, then
you have dx/dt, dy/dt
d distance/dt= sqrt (dx/dt ^2 + dy/dt ^2)
let the distance be z
z^2 = x^2 + y^2
z dz/dt = x dx/dt + y dy/dt
dx/dt = 4
dy/dt = 15/(2√(5x+1)) dx/dt
Now just plug in your numbers to find dz/dt
To find the rate of change of the distance from the particle to the origin, we need to determine the distance function first. Let's call the distance from the particle to the origin "d."
The distance formula is given by d = sqrt(x^2 + y^2), where x and y are the coordinates of the particle on the curve.
We're given that y = 3sqrt(5x) + 1. To find x, we need to solve the equation for x using the given point (3,12).
12 = 3sqrt(5x) + 1
Subtracting 1 from both sides, we get:
11 = 3sqrt(5x)
Dividing by 3, we have:
11/3 = sqrt(5x)
Squaring both sides, we get:
(11/3)^2 = 5x
121/9 = 5x
x = 121/45
Now that we have the x-coordinate of the particle, let's substitute it back into the equation of the curve to find the y-coordinate:
y = 3sqrt(5x) + 1
y = 3sqrt(5 * 121/45) + 1
y = 3sqrt(121/9) + 1
y = 3 * 11/3 + 1
y = 11 + 1
y = 12
So the coordinates of the particle are (121/45, 12).
Now, we can find the distance function by substituting these coordinates into the distance formula.
d = sqrt(x^2 + y^2)
d = sqrt((121/45)^2 + 12^2)
To find the rate of change of the distance with respect to time, we need to differentiate the distance function with respect to x and then multiply it by the rate at which x is changing.
Now we can find the derivative:
d/dx (d) = d/dx (sqrt((121/45)^2 + 12^2))
d(d)/dx = d/dx (sqrt((121/45)^2 + 144))
d(d)/dx = d/dx (sqrt(121^2/45^2 + 144))
d(d)/dx = d/dx (sqrt(14641/2025 + 144))
d(d)/dx = d/dx (sqrt(14641/2025 + 32400/2025))
d(d)/dx = d/dx (sqrt(47041/2025))
d(d)/dx = 1/2 * (47041/2025)^(-1/2) * (47041/2025)'
d(d)/dx = 1/2 * (47041/2025)^(-1/2) * (47041'(2025) - 2025'(47041))/(2025^2)
d(d)/dx = 1/2 * (47041/2025)^(-1/2) * (0 - 0)/(2025^2)
d(d)/dx = 0
Since d(d)/dx = 0, the rate of change of the distance from the particle to the origin is 0. This means that at the instant the particle passes through the point (3,12) with an x-coordinate increasing at a rate of 4 units per second, the distance from the particle to the origin remains constant.