how do i calculate the point on the curve when the ordinate changes 8 times faster than abscissa in the equation 6y=x^3+2
the conditions mean that the slope is 8.
y' is the slope, and since
6y' = 3x^2
y' = 1/2 x^2
If 1/2 x^2 = 8, then x = 4
So, at (4,11) the slope is 8
thank you
To calculate the point on the curve when the ordinate changes 8 times faster than the abscissa, we need to first understand the relationship between the abscissa (x) and the ordinate (y) in the given equation.
The equation you have provided is:
6y = x^3 + 2
We can rearrange this equation to express y in terms of x:
y = (1/6)x^3 + 2/6
y = (1/6)x^3 + 1/3
Now, let's consider the rate of change of the abscissa (x) and the ordinate (y). The rate of change is basically the derivative of the equation, representing how fast one variable changes with respect to the other.
To find the rate of change of the abscissa, we take the derivative of x with respect to y:
dx/dy = 1/(dy/dx)
Similarly, to find the rate of change of the ordinate, we take the derivative of y with respect to x:
dy/dx = (d/dx)((1/6)x^3 + 1/3)
Differentiating the equation, we get:
dy/dx = (1/6)(3x^2)
Given that the ordinate changes 8 times faster than the abscissa, we can set up the following equation:
dy/dx = 8(dx/dy)
Substituting the derivatives, we have:
(1/6)(3x^2) = 8(1/(dy/dx))
Simplifying further:
(1/6)(3x^2) = 8(dx/dy)
Now, we need to solve this equation to find the value of x. Let's continue with the calculation:
(1/6)(3x^2) = 8(dx/dy)
(1/6)(3x^2) = 8(1/(6x^2))
Multiplying both sides by 6x^2, we get:
3x^2 = 8
To solve for x, we divide both sides by 3:
x^2 = 8/3
Taking the square root of both sides, we find:
x = ± sqrt(8/3)
Therefore, the abscissa (x) can be either positive or negative square root of 8/3.
To find the corresponding ordinate (y), we substitute the value of x into the equation:
y = (1/6)x^3 + 1/3
By using the value of x obtained earlier, you can calculate the corresponding y coordinates for the point on the curve when the ordinate changes 8 times faster than the abscissa.