If an object is moving along the curve y=x^3, at what points is the y-coordinate changing 3 times more rapidly then the x-coordinate?
To find the points where the y-coordinate changes 3 times more rapidly than the x-coordinate, we need to determine the rate of change of the y-coordinate and the rate of change of the x-coordinate.
The rate of change of the y-coordinate (∆y/∆x) can be found by taking the derivative of the equation y = x^3 with respect to x.
dy/dx = 3x^2
The rate of change of the x-coordinate (∆x/∆t) is always equal to 1, as time (t) is not mentioned in the question.
Now, we can set up the equation to find the points where the y-coordinate changes 3 times more rapidly than the x-coordinate.
3(∆y/∆x) = (∆x/∆t)
Substituting the derivatives we found earlier:
3(3x^2) = 1
9x^2 = 1
Simplifying the equation:
x^2 = 1/9
Taking the square root of both sides:
x = ±1/3
So, the x-coordinates of the points where the y-coordinate changes 3 times more rapidly than the x-coordinate are x = 1/3 and x = -1/3.
To find the corresponding y-coordinates, we substitute these x-values back into the original equation:
For x = 1/3:
y = (1/3)^3 = 1/27
For x = -1/3:
y = (-1/3)^3 = -1/27
Therefore, the points where the y-coordinate changes 3 times more rapidly than the x-coordinate are (1/3, 1/27) and (-1/3, -1/27).
To determine at which points the y-coordinate is changing 3 times more rapidly than the x-coordinate, we need to calculate the derivatives of both coordinates and compare their rates of change.
First, let's find the derivatives of y and x with respect to some independent variable, usually denoted as t (representing time, for example):
dy/dt = d/dt(x^3)
To find dx/dt, we need to express x in terms of t. Since we are given the function y = x^3, we can rewrite it as:
y = (t^3)^3
y = t^9
Now, we can find dx/dt by differentiating y = t^9 with respect to t:
dx/dt = d/dt(t^9)
Taking the derivative of t^9, we get:
dx/dt = 9t^8
Now, we can compare the rates of change of the y-coordinate and the x-coordinate:
(dy/dt) / (dx/dt) = (d/dt(x^3)) / (d/dt(t^9))
= 3x^2 / (9t^8)
= x^2 / (3t^8)
According to the problem, the y-coordinate is changing 3 times more rapidly than the x-coordinate. Mathematically, this can be expressed as:
(dy/dt) = 3(dx/dt)
Substituting the derivatives we found earlier, we have:
3x^2 / (9t^8) = 3(9t^8)
x^2 / (3t^8) = 9t^8
Simplifying the equation:
x^2 = 27t^16
x = ±3t^8
Based on the equation we obtained, the x-coordinate changes 3 times more rapidly than the y-coordinate when x is equal to ±3t^8.
Therefore, at any points on the curve y = x^3 where x equals ±3t^8, the y-coordinate changes 3 times more rapidly than the x-coordinate.
y = 3x^2
dy/dt = 6x dx/dt
but dy/dt = 3 dx/dt
3dx/dt = 6x dx/dt
3 = 6x
x = 1/2
if x=1/2, y = (1/2)^3 = 1/8
It would happen at the point (1/2 , 1/8)