At what point on the curve y = 1 + √ 2x is the tangent line perpendicular to the line 6x + 2y = 1?

Answer 1

Solution: First I am assuming that the equation if the curve is y = 1 + √(2x) and not 1+ √2 (x) simply because the latter is actually an equation of a straight line and not a curve

The tangent line to the curve has opposite reciprocal slope to the perpendicular Line 6x+2y=1

Rewrite the equation in slope intercept form 2y = -6x +1 , y = (-6/2)x +(1/2) = -3x + 1/2.

The slope of this line is m = -3 and the slope of the tangent line perpendicular to it is m' = -1/m = -1/-3 = 1/3

Recall that the derivative of the equation of the curve gives you the slope of the tangent line at point (x , y) Find the derivative of the equation y = 1+√(2x) , y = 1+√2 x^1/2, and y' = 1/2(√2) x^{1/2 -1} = (√2/2) x^-1/2

y' = (√2/2)( 1/x^1/2) = √2 / 2√x

Now set y' = 1/3, √2 / 2√x = 1/3 and solve for x , 3√2 = 2√x , and √x = 3√2/ 2 , square both sides

(√x)² = ( 3√2/2)² , x = (9.2/4) = 9/2, so to find y substitute x = 9/2 into y = 1+√(2x) , y = 1+√[2(9/2)]

y = 1+√9 = 1+3 = 4 and the point of tangency ( x , y) is ( 9/2, 4)

Answer 2

Perpendicular lines have opposite reciprocal slopes.

So, find the slope of the of the line 6x + 2y = 1.

Then you can find the slope of the tangent line.

The derivative gives you the slope of the tangent line at points (x,y)

So take the derivative, set the derivative equal to the slope of the tangent line and solve for x