2.8 Related Rates

Formulas You May Need To Know

Related rate problems are differentiated with respect to time. So, every variable, except t is differentiated implicitly. Ex. Two rates that are related. Given y = x2 + 3, find dy/dt when x = 1, given that dx/dt = 2. y = x2 + 3 Now, when x = 1 and dx/dt = 2, we have

Assign symbols to all given quantities and Procedure For Solving Related Rate Problems Assign symbols to all given quantities and quantities to be determined. Make a sketch and label the quantities if feasible. Write an equation involving the variables whose rates of change either are given or are to be determined. Using the Chain Rule, implicitly differentiate both sides of the equation with respect to t. Substitute into the resulting equation all known values for the variables and their rates of change. Solve for the required rate of change.

First, a review problem: Consider a sphere of radius 10cm. If the radius changes 0.1cm (a very small amount) how much does the volume change? The volume would change by approximately .

Now, suppose that the radius is changing at an instantaneous rate of 0 Now, suppose that the radius is changing at an instantaneous rate of 0.1 cm/sec. (Possible if the sphere is a soap bubble or a balloon.) The sphere is growing at a rate of . Note: This is an exact answer, not an approximation like we got with the differential problems.

Water is draining from a cylindrical tank at 3 liters/second Water is draining from a cylindrical tank at 3 liters/second. How fast is the surface dropping? Find (We need a formula to relate V and h. ) (r is a constant.)

Steps for Related Rates Problems: 1. Draw a picture (sketch). 2. Write down known information. 3. Write down what you are looking for. 4. Write an equation to relate the variables. 5. Differentiate both sides with respect to t. 6. Evaluate.

Truck A travels east at 40 mi/hr. Truck B travels north at 30 mi/hr. Truck Problem: Truck A travels east at 40 mi/hr. Truck B travels north at 30 mi/hr. How fast is the distance between the trucks changing 6 minutes later? B A

p Truck Problem: Truck A travels east at 40 mi/hr. Truck B travels north at 30 mi/hr. How fast is the distance between the trucks changing 6 minutes later? B A p

Ex. A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate of 1 foot per second. When this radius is 4 ft., what rate is the total area A of the disturbed water increasing. Givens: Given equation: Differentiate:

An inflating balloon Air is being pumped into a spherical balloon at the rate of 4.5 in3 per second. Find the rate of change of the radius when the radius is 2 inches. Given: r = 2 in. Equation: Diff. & Solve:

The velocity of an airplane tracked by radar An airplane is flying at an elevation of 6 miles on a flight path that will take it directly over a radar tracking station. Let s represent the distance (in miles)between the radar station and the plane. If s is decreasing at a rate of 400 miles per hour when s is 10 miles, what is the velocity of the plane. s 6 x

Given: Find: Equation: Solve: x2 + 62 = s2 To find dx/dt, we must first find x when s = 10 Day 1

A fish is reeled in at a rate of 1 foot per second from a bridge 15 ft. above the water. At what rate is the angle between the line and the water changing when there is 25 ft. of line out? x 15 ft.

Given: Find: Equation: Solve: x = 25 ft. h = 15 ft.

Ex. A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius r of the outer ripple in increasing at a constant rate of 1 foot per second. When this radius is 4 ft., what rate is the total area A of the disturbed water increasing. An inflating balloon Air is being pumped into a spherical balloon at the rate of 4.5 in3 per minute. Find the rate of change of the radius when the radius is 2 inches.

Example Given Find when x = 3 Note: we must differentiate implicitly with respect to t

Example Now substitute in the things we know Find other values we need when x = 3, 32 + y2 = 25 and y = 4

Example Result

Guidelines for Related-Rate Problems Identify given quantities, quantities to be determined Make a sketch, label quantities Write equation involving variables Using Chain Rule, implicitly differentiate both sides of equation with respect to t After step 3, substitute known values, solve for required rate of change

R1 Electricity R2 The combined electrical resistance R of R1 and R2 connected in parallel is given by R1 and R2 are increasing at rates of 1 and 1.5 ohms per second respectively. At what rate is R changing when R1 = 50 and R2 = 75?

Draining Water Tank Radius = 20, Height = 40 The flow rate = 80 gallons/min What is the rate of change of the radius when the height = 12?

Draining Water Tank At this point in time the height is fixed Differentiate implicitly with respect to t, Substitute in known values Solve for dr/dt

Assignment Lesson 3.7 Page 187 Exercises 1 – 7 odd, 13 – 27 odd

Example #1 A ladder 10 feet long is resting against a wall. If the bottom of the ladder is sliding away from the wall at a rate of 1 foot per second, how fast is the top of the ladder moving down when the bottom of the ladder is 8 feet from the wall? First, draw the picture:

We have dx/dt is one foot per second. We want to find dy/dt. X and y are related by the Pythagorean Thereom Differentiate both sides of this equation with respect to t to get When x = 8 ft, we have Therefore The top of the ladder is sliding down (because of the negative sign in the result) at a rate of 4/3 feet per second.

Example #2 A man 6 ft tall walks with a speed of 8 ft per second away from a street light atop an 8 foot pole. How fast is the tip of his shadow moving along the ground when he is 100 feet from the light pole. 18 ft 6 ft z - x x z

Let x be the man’s distance from the pole and z be the distance of the tip of his shadow from the base of the pole. Even though x and z are functions of t, we do not attempt to obtain implicit formulas for either. We are given that dx/dt = 8 (ft/sec), and we want to find dz/dt when x = 100 (ft). We equate ratios of corresponding sides of the two similar triangles and find that z/18 = (z-x)/6 Thus 2z = 3x

Implicit differentiation now gives 2 dz/dt = 3 dx/dt We substitute dx/dt = 8 and find that (dz/dt = 3/2) * (dx/dt = 3/2) * (8) = 12 So the tip of the man’s shadow is moving at 12 ft per second.

Try Me! A ladder 25 ft long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at 3 ft/sec, how fast is the top of the ladder sliding down the wall, when the bottom is 15 ft from the wall?

Solution t = the number of seconds in time that has elapsed since the ladder started to slide down the wall. y = the number of feet in distance from the ground to the top of the ladder at t seconds. x = the number of feet in the distance from the bottom of the ladder to the wall at t seconds.

Because the bottom of the ladder is pulled horizontally away from the wall at 3 ft/sec, dx/dt = 3. We wish to find dy/dt when x = 15. From the Pythagorean Thereom, we have y^2 = 625 – x^2 Because x and y are functions of t, we differentiate both sides of equation one with respect to t and obtain 2y dy/dt = -2x dx/dt giving us dy/dt = -x/y dx/dt

When x = 15, it follows from equation one that y = 20. Because dx/dt = 3, we get from equation two: dy/dt = (-15/20) * 3 = -9/4 Therefore, the top of the ladder is sliding down the wall at the rate of 2 ¼ ft/sec when the bottom is 15 ft from the wall. The significance of the minus sign is that y is decreasing as t is increasing.

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