find the coordinates of the point on the curve xy = 10 in the first quadrant such that the normal line passes through the origin
Did you sketch the curve?
Did you notice the nice symmetry?
The normal clearly must be the line y = x
and must intersect at (√10 , √10)
so the equation of the line is y = x
To find the coordinates of the point on the curve xy = 10 in the first quadrant such that the normal line passes through the origin, we need to follow these steps:
Step 1: Understand the problem
First, let's understand the problem. We have a curve defined by the equation xy = 10, and we need to find a point on this curve in the first quadrant. Additionally, the normal line passing through this point should pass through the origin (0, 0).
Step 2: Solve for y in terms of x
From the given equation, xy = 10, we can rearrange it to solve for y in terms of x:
y = 10/x
Step 3: Calculate the derivative
To find the slope of the tangent line at any given point (x, y) on the curve, we need to calculate the derivative of the equation y = 10/x. Taking the derivative gives us:
dy/dx = -10/x^2
Step 4: Find the slope of the normal line
The normal line to the curve is perpendicular to the tangent line at the same point. Since the product of the slopes of two perpendicular lines is -1, the slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is:
m = -1 / (dy/dx) = -1 / (-10/x^2) = x^2 / 10
Step 5: Find the coordinates of the point
Since the normal line passes through the origin (0, 0), we can use the point-slope form of the equation of a line to find the equation of the normal line. It is given by:
y - y1 = m(x - x1)
Where (x1, y1) is the point on the curve.
Substituting the values, we get:
y - 0 = (x^2 / 10)(x - 0)
Simplifying, we have:
y = x^3 / 10
Now, we need to find the point on the curve where both the curve equation (xy = 10) and the equation of the normal line intersect. Substituting y in xy = 10 with the value of y from the equation of the normal line, we have:
x(x^3 / 10) = 10
Simplifying, we get:
x^4 = 100
x = ∛100 = 4.6416 (approx)
Substituting this value of x into the equation of the normal line, we can find y:
y = (4.6416)^3 / 10
y ≈ 4.0414
Hence, the point on the curve xy = 10 in the first quadrant, where the normal line passes through the origin, is approximately (4.6416, 4.0414).