Answer:
To find the coordinates of the points on the curve where the normals are perpendicular to the line 6y + x = 5, we need to find the points where the slopes of the curve and the line are negative reciprocals of each other.
Step 1: Find the slope of the line 6y + x = 5.
We need to rewrite the equation in slope-intercept form (y = mx + b):
6y + x = 5
6y = -x + 5
y = -(1/6)x + 5/6
The slope of the line is -1/6.
Step 2: Find the derivative of the curve y = x^3 - 3x^2 - 3x + 1.
To find the slope of the curve at any point (x, y), we take the derivative of the curve with respect to x:
y = x^3 - 3x^2 - 3x + 1
dy/dx = 3x^2 - 6x - 3
Step 3: Set the derivative equal to the negative reciprocal of the slope of the line:
3x^2 - 6x - 3 = -1/(-1/6)
3x^2 - 6x - 3 = 6
Step 4: Solve for x:
3x^2 - 6x - 3 - 6 = 0
3x^2 - 6x - 9 = 0
Step 5: Factor the quadratic equation:
3(x^2 - 2x - 3) = 0
3(x - 3)(x + 1) = 0
Now, set each factor to zero and solve for x:
x - 3 = 0 --> x = 3
x + 1 = 0 --> x = -1
Step 6: Find the corresponding y-coordinates using the original equation y = x^3 - 3x^2 - 3x + 1:
For x = 3:
y = 3^3 - 3(3)^2 - 3(3) + 1
y = 27 - 27 - 9 + 1
y = -8
For x = -1:
y = (-1)^3 - 3(-1)^2 - 3(-1) + 1
y = -1 - 3 + 3 + 1
y = 0
The two points on the curve where the normals are perpendicular to the line 6y + x = 5 are (3, -8) and (-1, 0).
Explanation: