Find the point on the curve y=2x+5 closest to the point (0,9).
I know to use the distance formula but I cannot get a correct answer
the distance z from (x,y) to (0,9) is
z = √(x^2 + (y-9)^2) = √(x^2 + (2x-4)^2)
= √(5x^2 - 16x + 16)
dz/dx = (5x-8)/√(5x^2 - 16x + 16)
so, the minimum distance is 4/√5 when x = 8/5
The point on the line is (8/5,41/5)
Or, using the normal, we know that the closest point on the curve is where the normal goes through (0,9)
The slope of the normal is -1/2, so the line through (0,9) is
y-9 = -1/2 x
y = -1/2 x + 9
That line intersects y=2x+5 at (8/5,41/5)
To find the point on the curve closest to a given point, you can use the distance formula. However, in this case, we can take a shortcut by finding the point on the curve that intersects the line perpendicular to it from the given point.
Let's break it down into steps:
Step 1: Given the point (0, 9) and the curve equation y = 2x + 5, we want to find the point on the curve closest to (0, 9).
Step 2: Determine the slope of the curve. The given equation is y = 2x + 5, which means the slope of the curve is 2.
Step 3: Calculate the slope of the line perpendicular to the curve. Since the slope of the curve is 2, the slope of the line perpendicular to it will be the negative reciprocal of 2, which is -1/2.
Step 4: Find the equation of the line perpendicular to the curve using the slope (-1/2) and the given point (0, 9). The equation of a line in point-slope form is y - y1 = m(x - x1), where (x1, y1) is the given point. Plugging in the values, we get:
y - 9 = (-1/2)(x - 0)
y - 9 = (-1/2)x
Simplifying, we get:
y = (-1/2)x + 9
Step 5: Find the point where the line and the curve intersect. We can do this by setting the equations of the curve and the line equal to each other:
2x + 5 = (-1/2)x + 9
Solving for x, we get:
2.5x = 4
x = 4/2.5 = 8/5 = 1.6
Substituting the value of x back into the equation of the curve, we find:
y = 2(1.6) + 5 = 3.2 + 5 = 8.2
The point on the curve closest to the given point (0, 9) is approximately (1.6, 8.2).