Thread: Implicit Differentiation

Hi, I just wanna double check my solution:

The equation of a circle with radius

r and center at the origin is

x^2+ y^2= r^2

(a) Use implicit differentiation to find the slope of a tangent line to the circle at some point (x, y)

2x + 2y dy/dx = 0 [since r will always be a constant...right?]
dy/dx = -x/y

(b) Use this result to find the equations of the tangent lines of the circle at the points whose x coordinate is x = r / sqrt(3)

y = +&- sqrt(r^2 - x^2)
y = r sqrt(2)
dy/dx = +&- 1 / sqrt(6)
using the points and slope, i can find the equation.

(c) Use the same result to show that the tangent line at any point on the circle is perpendicular to the radial line drawn from that point to the center of the circle

we know that dy/dx = -x/y is the slope of the tangent. In a circle, the slope of the radial line to the center = rise of run = y/x. Since, slopes are negative reciprocals; therefore, perpendicular.

Thanks for spending your time in double checking my answer

How did you get your y in b) . I've worked on it but get a different answer.

Well, y = ± sqrt(r^2 - x^2)
y = ± sqrt (3r^2 - r^2) [x= r/sqrt(3)]
y = ± r sqrt(2)

Is this what you did??

Not exactly, the answer i get is: y = ± r (sqrt(2)/sqrt(3))

Oh yeah, you are right!!

therefore, dy/dx got (b) = ± 1/sqrt(2)

Originally Posted by ninja

therefore, dy/dx got (b) = ± 1/sqrt(2)

yep