DERIVATIVES WITH RESPECT TO TIME RELATED RATES DERIVATIVES WITH RESPECT TO TIME

How do you take the derivative with respect to time when “time” is not a variable in the equation? Consider a circle that is growing on the coordinate plane: Growing Circle Animation An example of an equation of a circle similar to the one on the animation would be one like a circle with its center at the origin and radius of 2. (Notice, there is no “time” variable in this equation) x2 + y2 = 4 What we need to find is the rate of change of the x position over a period of time and/or the rate of change of the y position over a period of time. That is

The procedure: Similar to implicit differentiation – x2 + y2 = 4 Then, solve for whatever rate you are looking for. (Let’s say we want the rate of change of y with respect to time – dy/dt)

In each case find the derivative with respect to ‘t’. Then find dy/dt.

DERIVATIVES WITH RESPECT TO TIME RELATED RATES DERIVATIVES WITH RESPECT TO TIME

What is a related rate?

TABLE OF CONTENTS AREA AND VOLUME PYTHAGOREAN THEOREM AND SIMILARITY TRIGONOMETRY MISCELLANEOUS EQUATIONS

AREA AND VOLUME RELATED RATES

Example 1 Suppose a spherical balloon is inflated at the rate of 10 cubic inches per minute. How fast is the radius of the balloon increasing when the radius is 5 inches?

Ex 1: Answer Volume of a Sphere: Given: Find: when r = 5 inches

Example 2 (a) 2 inches? (b) 1 inch? A shrinking spherical balloon loses air at the rate of 1 cubic inch per minute. At what rate is its radius changing when the radius is (a) 2 inches? (b) 1 inch?

Ex 2: Answer Volume of a Sphere: Given: Find: when a) r = 2 inches b) r = 1 inch

Example 3 The area of a rectangle, whose length is twice its width, is increasing at the rate of Find the rate at which the length is increasing when the width is 5 cm.

Ex 3: Answer Area of a rectangle: Given: l = 2w Find: when w = 5 cm l = 10 cm

Example 4 Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 10 ft high?

Ex 4: Answer Volume of a Cone: Given: d = h or 2r = h Find: when h = 10 ft Eliminate ‘r’ from the equation and simplify

Ex 4: Answer (con’t) Substitute in the specific values and solve. Take the derivative Table of contents

Example 5 An inverted conical container has a height of 9 cm and a diameter of 6 cm. It is leaking water at a rate of 1 cubic centimeter per minute. Find the rate at which the water level h is dropping when h equals 3cm.

Ex 5: Answer Volume of a Cone: Given: Find: when h = 3 cm 9 Since the base radius is 3 and the height of the cone is 9, the radius of the water level will always be 1/3 of the height of the water. That is r = 1/3h

Ex 5: Answer (con’t) 3 Volume of a Cone: 9 Table of contents

PYTHAGOREAN THEOREM AND SIMILARITY

Example 6 A 13 meter long ladder leans against a a vertical wall. The base of the ladder is pulled away from the wall at a rate of 1 m/s. Find the rate at which the top of the ladder is falling when the base of the ladder is 5m away from the wall.

Ex 6: Answer Given: Length of ladder – 13 m Find: when x = 5 m y x Use Pythagorean Theorem to relate the sides of the triangle!

Ex 6: Answer (con’t) By the Pythagorean Thm: 13 y x Find ‘y’ when x = 5 using Pythagorean Thm.

Ex 7: A balloon and a bicycle A balloon is rising vertically above a level straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and balloon increasing 3 sec later?

Ex 7: Balloon and Bicycle - solution Given: rate of balloon rate of cyclist Find: when x = ? and y = ? Distance = rate * time s y x

Ex 7: Balloon and Bicycle - solution

Ex 8: The airplane problem- A highway patrol plane flies 3 mi above a level, straight road at a steady pace 120 mi/h. The pilot sees an oncoming car and with radar determines that at the instant the line of sight distance from plane to car is 5 mi, the line of sight distance is decreasing at the rate of 160 mi/h. Find the car’s speed along the highway.

Ex 8: Airplane - solution Given: rate of plane: when s=5: Find: rate of the car:

Ex 8: Airplane – solution(con’t) 3 5 s 3 (x+p) x + p

Ex 8: Airplane – solution(con’t) 3 (x+p)

Example 9 A 6 foot-tall man is walking straight away from a 15 ft-high streetlight. At what rate is his shadow lengthening when he is 20 ft away from the streetlight if he is walking away from the light at a rate of 4 ft/sec.

Ex 9: Answer Given: streetlight – 15 ft man – 6 ft Find: when x = 20 ft 15 6 x s Set up a proportion using the sides of the large triangle and the sides of the small triangle.

Ex 9: Answer (con’t) 15 6 x s Table of contents

RELATED RATES WITH TRIGONOMETRY

Example 10 A ferris wheel with a radius of 25 ft is revolving at the rate of 10 radians per minute. How fast is a passenger rising when the passenger is 15 ft higher than the center of the ferris wheel?

Ex 10: Answer Given: Radius – 25 ft Find: when y = 15 ft. 25 y 

Ex 10: Answer Find cos  when y = 15 ft 25 y 

Example 11 A baseball diamond is a square with sides 90 ft long. Suppose a baseball player is advancing from second to third base at a rate of 24 ft per second, and an umpire is standing on home plate. Let  be the angle between the third base line and the line of sight from the umpire to the runner. How fast is  changing when the runner is 30 ft from 3rd base?

Ex 11: Answer Given: Side length – 90 ft. Find: when x = 30 ft. x 90 

Ex 11: Answer (con’t) Solve equation for d/dt. Find cos  when x = 30: x 90  Table of contents

MISCELLANEOUS EQUATIONS

Example 12 An environmental study of a certain community indicates that there will be units of a harmful pollutant in the air when the population is p thousand. The population is currently 30,000 and is increasing at a rate of 2,000 per year. At what rate is the level of air pollution increasing?

Ex 12: Answer Given: Find: when p =30 thous

Example 13 It is estimated that the annual advertising revenue received by a certain news paper will be R(x) = 0.5x2 + 3x + 160 thousand dollars when its circulation is x thousand. The circulation of the paper is currently 10,000 and is increasing at a rate of 2,000 per year. At what rate will the annual advertising revenue be increasing with respect to time 2 years from now?

Ex 13: Answer Given: R(x) = 0.5x2 + 3x + 160 Find: Currently – x = 10 thousand in 2 years - x = 14 thousand

2007 – Form B – Question 3 The wind chill is the temperature, in degrees Fahrenheit (oF), a human feels based on the air temperature, in degrees Fahrenheit, and the wind velocity v, in miles per hour (mph). If the air temperature is 32oF, then the wind chill is given by W(v) = 55.6 – 22.1v0.16 and is valid for 5 ≤ v ≤ 60 . a) Find W’(20). Using correct units, explain the meaning of W’(20) in terms of the wind chill. b) Find the average rate of change of W over the interval . Find the value of v at which the instantaneous rate of change of W is equal to the average rate of change of W over the interval . c) Over the time interval 0 ≤ t ≤ 4 hours, the air temperature is a constant 32oF. At time t = 0, the wind velocity is v = 20 mph. If the wind velocity increases at a constant rate of 5 mph, what is the rate of change of the wind chill with respect to time at t = 3 hours? Indicate units of measure.