Slope of a Line Tangent to a Circle – Implicit Version | Lecture Note - Edubirdie

University:
Massachusetts Institute of Technology
Course:
18.01SC | Single Variable Calculus
Academic year:

2021
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399

Pages:

1
Author:

sperrtrab7z

Slope of a line tangent to a circle – implicit version We just ﬁnished calculating the slope of the line tangent to a point (x, y) on the top half of the unit circle. In this calculation we started by solving the equation x2 + y 2 = 1 for y, chose one “branch” of the solution to work with, then used the chain rule, the power rule and some algebra of exponents to compute the dy = − xy . derivative dx We’ll now see how we could have used implicit diﬀerentiation to do the same calculation much more easily. In fact, we’ll ﬁnd the slope of a line tangent to any point on the unit circle. d We don’t need to solve for y — we can just apply the operator dx to both sides of the original equation: x2 + y 2 � d � 2 x + y2 dx d 2 d 2 (x ) + (y ) dx dx = 1 d (1) dx = = 0 We can easily take the derivative of the ﬁrst term. For the second term, applying the chain rule with the inside function y and outside function u2 gives us: dy dx dy 2y dx dy dx 2x + 2y = 0 = −2x = − x y This is the same answer, but we didn’t have to restrict ourselves to just the top half of the circle or use any square roots. Implicit diﬀerentiation made this calculation much easier.

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