Posted by **Christine** on Sunday, August 2, 2009 at 7:21pm.

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).

- Math -
**MathMate**, Sunday, August 2, 2009 at 9:59pm
(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

- Math -
**deon**, Tuesday, April 5, 2011 at 5:16pm
slope of -8y-3x+5=10x

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