Question

a) determine the equation of the line that is perpendicular to the tangent to y=5x^2 at (1,5).

b) determine the equation of the line that passes through (2,2) and is parallel to the line tangent to y=-3x^3-2x at (-1,5).

Answer 1

(a)
y=5x²
dy/dx=5*2x=10x, slope of tangent
slope of line perpendicular to tangent
= -1/(10x)
At (1,5), slope of perpendicular line
= -1/(10*1) = -1/10
Line passing through (1,5)
(y-5)=-(1/10)(x-1)
y =-(1/10)x + 5.10

Test for perpendicularity at (1,5)
-(1/10) * (10*1)
= -1 Line and tangent are perpendicular.

(b) y=-3x^3-2x at (-1,5).
dy/dx = -9x² -2 = -9-2 = -11 at (-1,5)
Slope of tangent = -11
Line passing through (2,2) with slope of -11
(y-2) = -11(x-2)
y-2 = -11x + 22
y = -11x + 24

Answer 2

slope of -8y-3x+5=10x