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