The slope of the tangent to a curve at any point (x, y) on the curve is -x/y . Find the equation of the curve if the point (3,-4) on the curve.
you might have noticed that if we find dy/dx for a circle of the form x^2 + y^2 = r^2 we get
2x + 2y dy/dx = 0
dy/dx = -2x/(2y) = -x/y
So we had a circle of the form x^2 + y^2 = r^2
plug in the given point to find r^2
and you are done.
To find the equation of the curve, we need to integrate the given information.
Let's start by integrating the given slope function with respect to x:
∫(-x/y) dx = ∫-xdx / y
To integrate -x with respect to x, we use the power rule: integrate x^n into (1/(n+1))x^(n+1). In this case, n = 1.
∫-xdx = (-1/2) x^2 + C
Now, we can rewrite the integral as:
(-1/2) x^2 / y + C = (-x^2)/(2y) + C
To find the equation of the curve, we need to solve for C. We can use the given point on the curve (3, -4) to determine the value of C.
Substituting the coordinates into the equation:
(-3^2)/(2(-4)) + C = -9/8 + C
C = -9/8 + 9/8 = 0 (after simplification)
Therefore, the equation of the curve is:
(-x^2)/(2y) + 0 = (-x^2)/(2y) = (-x^2)/(2y)