Study Notes on Implicit Differentiation and Related Rates

Implicit Differentiation and the Chain Rule

Introduced implicit differentiation in relation to the chain rule.
The derivative of a variable can be expressed in terms of other derivatives by the chain rule.
- Example: As the expression for the derivative dy/dx is multiplied by dx/dt, results in cancellation of dx.
- Remaining expression is dy/dt = (dy/dx)(dx/dt).

Given a scenario where dy/dt = -4 cm/s (decreasing).
- This requires accounting for the negative sign to indicate that the distance is decreasing.
The calculation includes:
- -4 = (1/2)(1/sqrt(1 + 2^2))(3)(2^2)(dx/dt).
Final result would isolate dx/dt, calculated as -2 cm/s.

Situation described with a baseball diamond having sides of 90 feet.
- Player runs from home to first, second, and third bases.
Problem focuses on the distance between the ball (running along the third base line) and first base.
- Situation emphasizes the importance of defining variables.
Define Variables:
- b = Distance the ball has traveled from home plate.
- f = Distance from the ball to first base (this distance is changing).
- Fixed distance between home plate and first base = 90 ft.
- At halfway to third base, b = 45 ft.
Geometrically relates to Pythagorean theorem:
- b^2 + 90^2 = f^2.
- Here, d(b)/dt = 100 ft/s (speed of the ball).
Taking derivatives for rates of change:
- Apply chain rule:
- 2b(d(b)/dt) + 0 = 2f(d(f)/dt).
Isolating d(f)/dt:
- d(f)/dt = (b /f)(d(b)/dt).
Substituting values to find rate at which distance between the ball and first base changes when b = 45.
- At this moment, f is calculated as f = sqrt(45^2 + 90^2) = 45sqrt(5).
- Final formula yields d(f)/dt = (100/√5).
Units: Distance (ft) and time (seconds).

If the batter has run an eighth of the way to first base, the new speed is 30 ft/s.
This requires recalculating lengths using new variables and solidifying a similar approach with three variables instead of two.
Important relationship still follows the Pythagorean theorem but introduces one more variable to consider (distance from ball to runner).

Water is pumped into a cone-shaped tank at a rate of 15 cubic feet per minute.
- Tank dimensions: 10 ft deep, top radius = 5 ft.
- Question focuses on the rate at which water level is rising when depth (h) = 4 ft.
Define Variables:
- r = radius of water surface, h = height of water.
- dV/dt = rate of water pumped in = +15 ft³/min.
Volume formula for a cone:
- V = (1/3)πr²h.
Rate of change (dV/dt) involves product and chain rule because both r and h are changing over time.
- Relationships established through ratios from the geometry of similar triangles between larger and smaller cones provide a substitute for r.
After differentiating and isolating d(h)/dt, the relationship yields specific values that can be plugged into findings to compute rates.
- d(h)/dt = 15/(4π) ft/min.

When introducing a leak in the tank while water is pumped, a new rate of change expression emerges: dV/dt = rate water in - rate water leaking.
- Calculation using volume formula still applies with adjustments for the leaking rate.
If the height of water is rising at only 1 ft/min instead of computed level (when h = 4), the new rate of leak can be derived from changes in the volume rate.
- 15 - dV/dt = leaking rate expression yields comparative analysis against previous values.

Emphasis on understanding change in systems through defined rates and relationships between various parameters.
Critical application of the chain and product rules within related rates problems, particularly in geometric contexts such as the baseball diamond and varying volume of a cone.