Implicit Differentiation Related Rates

Implicit DifferentiationExplicit form: y = f(x).Implicit form: F (x, y) = 0 - sometimes you cannot solve for y. Example: circle.How to find the derivative y ′ of implicit function:1. Differentiate both sides of an equation. Must use the chain rule for all terms with y.2. Solve for y ′ .Practice Problems.1. Find the derivative y ′ .a) x 2 + xy = 6b) x 3 + 12xy = y 3c) xe y + x 2 = y 22. Find an equation of the line tangent to the graph of the given equation at theindicated point.a) x 2 + y 2 = 13; (3, 2)b) x ln y = 2x 3 − 2y; (1, 1)c) x 2 + y 2 = e y ; (1, 0)Solutions.1. a) y ′ = −(2x+y)/x b) y ′ = (x 2 +4y)/(y 2 −4x) c) y ′ = (e y +2x)/(2y −xe y )2. a) y = −3/2x + 13/2 b) y = 2x − 1 c) y = 2x − 2Related RatesTo solve a related rates problem, follow the steps:1. Sketch a diagram if possible.2. Write down all the variables along with the rates given.3. Write down the equation that relates all the variables.4. Differentiate the equation implicitly.5. Solve for the unknown rate.

Practice Problems.1. Suppose a spherical balloon is inflated at the rate of 10 cubic centimeters per minute.How fast is the radius of the balloon increasing at the time when the radius is 5cm? Recall that the formula for the volume of a sphere is V = 4/3 πr 3 .2. A 20-foot ladder is leaning against the wall. If the base of the ladder is sliding awayfrom the wall at the rate of 3 feet per second, find the rate at which the top of theladder is sliding down when the top of the ladder is 8 feet from the ground.3. Water leaking onto a floor creates a circular puddle with an area that increases atthe rate of 3 square centimeters per minute. How fast is the radius of the puddleincreasing when the radius is 10 cm? Recall that the formula for the area of a circleis A = r 2 π.4. Pat walks at the rate of 5 feet per second towards a 24-feet-tall street lamp. If Patis 6 feet tall, how fast is the tip of Pat’s shadow moves along the ground.5. Assume that the number of bass in the pond is related to the level of polychlorinatedbiphenyls (PCBs, a group of industrial chemicals used in plasticizers, fire retardantsand other materials) in the pond. The bass population is modeled byy = 25001 + xwhere x represents the PCB level in parts per million (ppm) and y represents thenumber of bass in the pond. If the level of PCBs is increasing at the rate of 40 ppmper year, find the rate at which is the number of bass changing when there are 100bass in the pond.Solutions: 1. 1/(10π) cm per min. 2. −6.87 ft per sec. 3. 3/(20π) cm permin. 4. −5/3 ft per sec. 5. −160 bass per year