I need help with this problem: At what value of x do the graphs of y=x^(2) and y=-sqrt (x) have perpendicular tangent lines?

Question

How do I tackle this problem? Is this another word for the normal line?

Answer 1

This isn't another word for the normal line because both lines that you're talking about are still tangents to each curve, and even though they'll cross the curves at the same x-value, the y-values on the curves at that x value won't (necessarily) be the same. The tangent lines just happen to have slopes that make them perpendicular to each other. The key fact is: If m₁ and m₂ are slopes of perpendicular lines, then m₁ * m₂ = -1. In other words, m₁ = -1/m₂ 

The general idea is: find the derivative of each function (which is really just a way of finding the slope for a particular x-value). Multiply the derivatives together, set them equal to -1 and solve for x.

 

You can double-check any solutions by plugging the value(s) into the derivatives and making sure the derivatives exist and are negative reciprocals. 

 

Answer 2

As Kathleen said, take the derivatives of the functions,

multiply them together, set equal to -1, solve for x:

 

y = x²

y' = 2x

 

y = -x^1/2

y' = (-1/2)x^-1/2 = -1/(2x^1/2)

 

Multiply the derivatives together and set = -1

 

2x[ -1/(2x^1/2)] = -1

-2x/2x^1/2 = -1

-x^1/2 = -1

x^1/2 = 1

x = 1

 

The graphs have perpendicular tangent lines at x=1

 

Verify that the slopes ... i.e. the derivatives of the functions ...

of the tangent lines are negative reciprocals at x=1

 

y' = 2x = 2(1) = 2

 

y' = -1/2x^1/2 = -1/2(1)^1/2 = -1/2

 

The slopes are negative reciprocals