Question
The slope of the line normal to the curve e^x − x^3 + y^2 =10 at the point (0, 3) is ____
Answers
Reiny
take derivative first
e^x - 3x^2 + 2y dy/dx = 0
dy/dx = (3x^2 - e^x)/2y
at (0,3)
dy/dx = (0 - 1)/6 = -1/6
so the slope of the tangent at the given point is -1/6
thus the slope of the normal at that point is +6
e^x - 3x^2 + 2y dy/dx = 0
dy/dx = (3x^2 - e^x)/2y
at (0,3)
dy/dx = (0 - 1)/6 = -1/6
so the slope of the tangent at the given point is -1/6
thus the slope of the normal at that point is +6
Molly
Positive 6 was correct!