The slope of the tangent line to a curve at any point (x, y) on the curve is x/y. What is the equation of the curve if (4, 1) is a point on the curve?
x2 − y2 = 15
x2 + y2 = 15
x + y = 15 <- my answer
xy = 15
to get an x and y in the slope, because that was the result of a derivative, you must have had an equation
x^2=y^2 + k or
x^2-y^2=k
proof
2x-2y dy/dx=d/dx K=0
dy/dx=x/y
to get an x and y in the slope, because that was the result of a derivative, you must have had an equation
x^2=y^2 + k or
x^2-y^2=k
proof
2x-2y dy/dx=d/dx K=0
dy/dx=x/y
opps, you want the k for the funcion.
4^2-1^2=k
k=15
x^2-y^2=15
To find the equation of the curve, we need to use the information given about the slope of the tangent line at any point (x, y) on the curve.
The slope of the tangent line is given as x/y. This means that the derivative of the curve with respect to x is equal to x/y.
To find the derivative, we differentiate the equation of the curve with respect to x. Let's say the equation of the curve is y = f(x).
Differentiating y = f(x) with respect to x gives us dy/dx. Thus, we have:
dy/dx = x/y
Now, let's integrate both sides of this equation to eliminate the derivative. Integrating dy/dx with respect to x gives us y, and integrating x/y with respect to x gives us ln|y| + C, where C is the constant of integration.
So, we have:
y = ln|y| + C
To find the constant of integration C, we can use the given point (4, 1) on the curve. Substituting x = 4 and y = 1 into the equation, we get:
1 = ln|1| + C
1 = 0 + C
C = 1
Therefore, the equation of the curve is:
y = ln|y| + 1
This equation represents the curve with the given point (4, 1). None of the answer choices x2 − y2 = 15, x2 + y2 = 15, x + y = 15, or xy = 15 match the equation of the curve.