find an equation of the normal line to the curve y=radx-3 that is parallel to the line 6x+3y-4=0
actually, the normal line is
y-1 = -2(x-4)
y = -2x+9
To find the equation of the normal line to the curve y = rad(x) - 3 that is parallel to the line 6x + 3y - 4 = 0, we need to follow several steps:
Step 1: Identify the slope of the given line
To determine the slope of the line 6x + 3y - 4 = 0, we need to rearrange the equation in the slope-intercept form (y = mx + b), where m represents the slope.
Rearranging the equation, we have:
3y = -6x + 4
y = -2x + 4/3
From this equation, we can conclude that the slope of the given line is -2.
Step 2: Determine the slope of the normal line
Since the normal line is perpendicular to the curve at a given point, the slope of the normal line is equal to the negative reciprocal of the derivative of the curve at that point.
To find the derivative d(y)/dx of y = rad(x) - 3, we differentiate the equation with respect to x:
d(y)/dx = d(rad(x))/dx - d(3)/dx = (1/2)*(1/sqrt(x))
We can rewrite the equation as:
d(y)/dx = 1/(2*sqrt(x))
Step 3: Find the slope of the normal line at a given point on the curve
To find the slope of the normal line at a specific point on the curve, we substitute the x-coordinate of the point into the derivative function.
Since we aim to find a normal line parallel to the line 6x + 3y - 4 = 0, we need to consider at what point the curve intersects this line. To find the point of intersection, we solve the system of equations formed by y = rad(x) - 3 and 6x + 3y - 4 = 0.
5x + 3(rad(x) - 3) - 4 = 0
At this point, we need to solve for x numerically, which involves using numerical methods or a graphing calculator. Let's assume that this equation has a solution of x = a.
Step 4: Find the y-coordinate of the point of intersection
Using the value of x = a obtained from step 3, we can plug it back into the equation y = rad(x) - 3 to find the corresponding y-coordinate.
Therefore, the coordinates of the point of intersection are (a, rad(a) - 3).
Step 5: Calculate the slope of the normal line at the point of intersection
Substitute the x-coordinate of the point of intersection into the derivative function:
Slope = 1/(2*sqrt(a))
Step 6: Find the equation of the normal line
We now have the slope of the normal line (from step 5) and a point on the line (from step 4). We can use the point-slope form of a line to find the equation.
Using the point-slope form (y - y1) = m(x - x1), we substitute the values:
(y - (rad(a) - 3)) = (1/(2*sqrt(a))) * (x - a)
Simplifying this equation, we can rearrange it into the standard form (Ax + By + C = 0):
-2*sqrt(a)*x + 2*rad(a)*x - a - 2*sqrt(a)*y + 2*rad(a) - 6 = 0
Thus, the equation of the normal line to the curve y = rad(x) - 3, which is parallel to the line 6x + 3y - 4 = 0, is:
-2*sqrt(a)*x + 2*rad(a)*x - a - 2*sqrt(a)*y + 2*rad(a) - 6 = 0, where a is the x-coordinate of the point of intersection.
Let the point of contact be P(a,b)
or P(a, √(a-3) )
If it is to be a normal at P, then the tangent at P must be perpendicular to 6x + 3y - 4 = 0
that is,
the tangent must have a slope of +1/2
dy/dx = (1/2)(x-3)^(-1/2)
= 1/( 2√(x-3) )
1/(2√(x-3) ) = 1/2
2√(x-3) = 2
√(x-3) = 1
x-3 = 1
x = 4
so point P is (4,1)
so the normal must have a slope of -2
equation:
y = -2x + b, with (4,1) on it ....
4 = -2(1) + b
b = 6
normal equation : y = -2x + 6
checking by looking at graph
http://www.wolframalpha.com/input/?i=+plot+y+%3D+√%28x-3%29%2C+y+%3D+-2x%2B6%2C+6x+%2B+3y+-+4+%3D+0