Find the xy-equation of the curve that passes through (-2, -2) and whose slope at any point on the curve is equal to 5 times the x-coordinate of that point
y'= 5x
y= 2.5 x^2 + b
-2=2.5*(4)+b
B= 12
y= 2.5 x^2 + 12
To find the equation of a curve that passes through a given point and has a slope equal to a given function of the x-coordinate, we can use the technique of separation of variables and integration.
Let's start by denoting the equation of the curve as y = f(x). Given that the slope at any point on the curve is equal to 5 times the x-coordinate of that point, we can write the derivative of y with respect to x as dy/dx = 5x.
To find the equation of the curve, we need to integrate both sides of this equation. Integrating dy/dx = 5x with respect to x gives us ∫dy = ∫5x*dx.
Integrating ∫dy will yield y, and integrating ∫5x*dx with respect to x will yield (5/2)x^2 + C, where C is the constant of integration.
Therefore, the equation of the curve is y = (5/2)x^2 + C.
We can now use the given point (-2, -2) to find the value of the constant C. Substituting the x and y values into the equation, we have -2 = (5/2)(-2)^2 + C.
Simplifying this equation, we get -2 = (5/2) * 4 + C, which can be further simplified as -2 = 10 + C.
Finally, we find that C = -12.
The xy-equation of the curve that passes through (-2, -2) and whose slope at any point on the curve is equal to 5 times the x-coordinate of that point is:
y = (5/2)x^2 - 12.