Lesson 9: Implicit Differentiation & Related Rates Intro

1. Implicit Differentiation

Sometimes y is defined implicitly (not solved for y), e.g. \(x^2 + y^2 = 25\) (circle).

Differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\):

\(2x + 2y dy/dx = 0\)

\(dy/dx = -x/y\) (slope of tangent at any point on circle)

Very useful when \(y\) cannot be easily isolated.

Exercise 1

For \(x^2 + y^2 = 100\), find \(dy/dx\):
\(2x + 2y dy/dx = 0 \rightarrow dy/dx = \) /

2. Related Rates – Introduction

When quantities change over time, their rates are related.

Example: circle radius \(r\) increases at \(dr/dt = 2\) cm/s. How fast is area increasing when r = 5 cm?

\(A = \pi r^2\)

\(dA/dt = 2\pi r dr/dt\)

When \(r = 5,\, dA/dt = 2\pi·5·2 = 20\pi\) cm²/s

Steps: relate variables → differentiate both sides w.r.t. time → plug in known rates.

Exercise 2

Volume of sphere \(V = (4/3)\pi r^3\)
If \(dr/dt = 3\) cm/s, find \(dV/dt\) when \(r = 4\) cm:
\(dV/dt = 4\pi r^2 ·\) \(=\) \(\pi \) cm³/s

Exercise 3

Which situations use related rates? (Select all that apply)
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