• 6.2 Related Ratesap Calculus Worksheet
• 6.2 Related Ratesap Calculus Pdf
• 6.2 Related Ratesap Calculus Solutions
• 6.2 Related Ratesap Calculus Answers

AP Calculus LHopitals Rule and Related rates. AP Calculus Learning Objectives Explored in this Section. Calculate derivatives; Interpret the meaning of a derivative.

6.2: Related Rates

• This lesson contains the following Essential Knowledge (EK) concepts for the.AP Calculus course.Click here for an overview of all the EK's in this course. EK 2.3C2 EK 2.3D1.
• Check out all of my Calculus Videos and Notes at: http://wowmath.org/Calculus/CalculusNotes.html.

• You now know how to take the derivative with respect to the independent variable. In other words, you know how to determine a function's rate of change when given the input's rate of change. But what if the independent variable was itself a function? What if, for example, the input was a function of time? How do we identify how the function changes as time changes? This subunit will explore the answers to these questions.
  

• Read Section 6.2 (pages 127-132). Another application of the chain rule, related rates problems apply to situations where multiple dependent variables are changing with respect to the same independent variable. Make note of the summary in the middle of page 128.
  

• Watch this video until 40:30. The majority of the lecture is about related rates, despite its title.
  

• Work through problems 1, 3, 5, 11, 14, 16, 19-21, and 25 for Exercise 6.2. When you are done, check your answers against Appendix A.
  

Suppose we have two variables $x$ and $y$ (in most problems theletters will be different, but for now let's use $x$ and $y$) whichare both changing with time. A 'related rates' problem is a problemin which we know one of the rates of change at a given instant—say,$ds dot x = dx/dt$—and we want to find the other rate $ds dot y = dy/dt$ at thatinstant. (The use of $ds dot x$ to mean $dx/dt$ goes back to Newton andis still used for this purpose, especially by physicists.)

If $y$ is written in terms of $x$, i.e., $y=f(x)$, then this is easyto do using the chain rule:$$dot y = {dyover dt}={dyover dx}cdot{dxover dt}={dyover dx}dot x.$$That is, find the derivative of $f(x)$, plug in the value of$x$ at the instant in question, and multiply by the given value of$ds dot{x}=dx/dt$ to get $ds dot{y}=dy/dt$.

6.2 Related Ratesap Calculus Worksheet

Example 6.2.1 Suppose an object is moving along a path described by $ds y=x^2$, thatis, it is moving on a parabolic path. At a particular time, say $t=5$,the $x$ coordinate is 6 and we measure the speed at which the $x$ coordinate of the object ischanging and find that $dx/dt = 3$. At the same time, how fast is the$y$ coordinate changing?

Using the chain rule, $ds dy/dt = 2xcdot dx/dt$. At $t=5$ we know that$x=6$ and $dx/dt=3$, so $dy/dt = 2cdot 6cdot 3 = 36$.

In many cases, particularly interesting ones,$x$ and $y$ will be related in some other way, for example$x=f(y)$, or $F(x,y)=k$, or perhaps $F(x,y)=G(x,y)$, where $F(x,y)$and $G(x,y)$ are expressions involving both variables. In all cases, youcan solve the related rates problem by taking the derivative of both sides,plugging in all the known values (namely, $x$, $y$, and $ds dot{x}$), andthen solving for $ds dot{y}$.

To summarize, here are the steps in doing a related rates problem:

1. Decide what the two variables are.

2. Find an equation relating them.

3. Take $d/dt$ of both sides.

4. Plug in all known values at the instant in question.

5. Solve for the unknown rate.

Example 6.2.2 A plane is flying directly away from you at 500 mph at an altitude of3 miles. How fast is the plane's distance from you increasing at themoment when the plane is flying over a point on the ground 4 milesfrom you?

To see what's going on, we first draw a schematic representation ofthe situation, as in figure 6.2.1.

6.2 Related Ratesap Calculus Pdf

Because the plane is in level flight directly away from you, the rateat which $x$ changes is the speed of the plane, $dx/dt=500$. Thedistance between you and the plane is $y$; it is $dy/dt$ that we wishto know. By the Pythagorean Theorem we know that $ds x^2+9=y^2$. Takingthe derivative:$$ 2x dot x = 2ydot y.$$We are interested in the time at which $x=4$; at this time we knowthat $ds 4^2+9=y^2$, so $y=5$. Putting together all the information weget$$2(4)(500)=2(5)dot y.$$Thus, $ds dot y=400$ mph.

Example 6.2.3 You are inflating a spherical balloon at the rate of 7 cm${}^3$/sec. Howfast is its radius increasing when the radius is 4 cm?

6.2 Related Ratesap Calculus Solutions

Here the variables are the radius $r$ and the volume $V$. We know $dV/dt$,and we want $dr/dt$. The two variables are related by means of theequation $ds V=4pi r^3/3$. Taking the derivative of both sides gives$ds dV/dt=4pi r^2dot r$. We now substitute the values we know at theinstant in question: $ds 7=4pi 4^2dot r$, so$ds dot r=7/(64pi)$ cm/sec.

Example 6.2.4 Water is poured into a conical container at the rate of 10cm${}^3$/sec. The cone points directly down, and it has a height of30 cm and a base radius of 10 cm; see figure 6.2.2.How fast is the water level rising when the water is 4 cm deep (at itsdeepest point)?

The water forms a conical shape within the big cone; itsheight and base radius and volume are all increasingas water is poured into the container. This means that we actually havethree things varying with time: the water level $h$ (the height of the coneof water), the radius $r$ of the circular top surface of water (the baseradius of the cone of water), and the volume of water $V$. The volume of a cone is given by$ds V=pi r^2h/3$. We know $dV/dt$, and we want $dh/dt$. Atfirst something seems to be wrong: we have a third variable $r$ whose ratewe don't know.

But the dimensions of the cone of water must have the sameproportions as those of the container. That is, because of similar triangles, $r/h=10/30$ so $r=h/3$. Now we can eliminate $r$ from theproblem entirely: $ds V=pi(h/3)^2h/3=pi h^3/27$. We takethe derivative of both sides and plug in $h=4$ and $dV/dt=10$, obtaining$ds 10=(3picdot 4^2/27)(dh/dt)$. Thus, $dh/dt=90/(16pi)$cm/sec.

Example 6.2.5 A swing consists of a board at the end of a 10 ft long rope. Think of theboard as a point $P$ at the end of the rope, and let $Q$ be the point ofattachment at the other end. Suppose that the swing is directly below $Q$at time $t=0$, and is being pushed by someone who walks at 6ft/sec from left to right. Find (a) how fast the swing is rising after 1sec; (b) the angular speed of the rope in deg/sec after 1 sec.

We start out by asking: What is the geometricquantity whose rate of change we know, and what is the geometric quantitywhose rate of change we're being asked about? Note that the person pushingthe swing is moving horizontally at a rate we know. In other words,the horizontal coordinate of $P$ is increasing at 6 ft/sec. In the$xy$-plane let us make the convenient choice of putting the origin at thelocation of $P$ at time $t=0$, i.e., a distance 10 directly below the pointof attachment. Then the rate we know is $dx/dt$, and in part (a) the rate we want is $dy/dt$ (the rate at which $P$ is rising). In part(b) the rate we want is $ds dot{theta}=dtheta/dt$, where $theta$ stands forthe angle in radians through which the swing has swung from the vertical.(Actually, since we want our answer in deg/sec, at the end we must convert$dtheta/dt$ from rad/sec by multiplying by $180/pi$.)

(a) From the diagram we see that we have a right triangle whose legsare $x$ and $10-y$, and whose hypotenuse is 10. Hence$ds x^2+(10-y)^2=100$. Taking the derivative of both sides we obtain:$2xdot{x}+2(10-y)(0-dot{y})=0$. We now look at what we know after 1second, namely $x=6$ (because $x$ started at 0 and has been increasing atthe rate of 6 ft/sec for 1 sec), $y=2$ (because we get $10-y=8$ fromthe Pythagorean theorem applied to the triangle with hypotenuse 10 andleg 6), and $ds dot{x}=6$. Putting in these values gives us$2cdot 6cdot 6-2cdot 8dot{y}=0$, from which we can easily solvefor $ds dot{y}$: $ds dot{y}=4.5$ ft/sec.

(b) Here our two variables are $x$ and $theta$, so we want to use thesame right triangle as in part (a), but this time relate $theta$ to$x$. Since the hypotenuse is constant (equal to 10), the best way todo this is to use the sine: $sintheta=x/10$. Taking derivatives weobtain $ds (costheta)dot{theta}=0.1dot{x}$. At the instant inquestion ($t=1$ sec), when we have a right triangle with sides6–8–10, $ds costheta=8/10$ and $ds dot{x}=6$. Thus$(8/10)dot{theta}=6/10$, i.e., $ds dot{theta}=6/8=3/4$ rad/sec, orapproximately $43$ deg/sec.

We have seen that sometimes there are apparently more than twovariables that change with time, but in reality there are just two, asthe others can be expressed in terms of just two. But sometimes therereally are several variables that change with time; as long as youknow the rates of change of all but one of them you can find the rateof change of the remaining one. As in the case when there are just twovariables, take the derivative of both sides of the equation relating all ofthe variables, and then substitute all of the known values and solve forthe unknown rate.

Example 6.2.6 A road running north to south crosses a road going east to west at thepoint $P$. Car A is driving north along the first road, and car B isdriving east along the second road. At a particular time car A is $10$kilometers to the north of $P$ and traveling at 80 km/hr, while car Bis 15 kilometers to the east of $P$ and traveling at 100 km/hr.How fast is the distance between the two carschanging?

6.2 Related Ratesap Calculus Answers

Let $a(t)$ be the distance of car A north of $P$ at time $t$, and$b(t)$ the distance of car B east of $P$ at time $t$, and let $c(t)$be the distance from car A to car B at time $t$. By the PythagoreanTheorem, $ds c(t)^2=a(t)^2+b(t)^2$. Taking derivativeswe get $ds 2c(t)c'(t)=2a(t)a'(t)+2b(t)b'(t)$, so$$ dot{c}={adot{a}+bdot{b}over c}={adot{a}+bdot{b}over sqrt{a^2+b^2}}.$$Substituting known values we get:$$dot{c}={10cdot 80+15cdot100over sqrt{10^2+15^2}}={460oversqrt{13}} approx 127.6 hbox{km/hr}$$at the time of interest.

Notice how this problem differs from example 6.2.2. In both cases we started with the Pythagorean Theorem andtook derivatives on both sides. However, inexample 6.2.2 one of the sides was a constant(the altitude of the plane), and so the derivative of the square ofthat side of the triangle was simply zero. In this example, on theother hand, all three sides of the right triangle are variables, eventhough we are interested in a specific value of each side of thetriangle (namely, when the sides have lengths 10 and 15). Make sure thatyou understand at the start of the problem what are the variables andwhat are the constants.

Exercises 6.2

Ex 6.2.1A cylindrical tank standing upright (with one circular base on theground) has radius 20 cm. How fast does the water level in thetank drop when the water is being drained at 25 cm${}^3$/sec?(answer)

Ex 6.2.2A cylindrical tank standing upright (with one circular base on theground) has radius 1 meter. How fast does the water level in thetank drop when the water is being drained at 3 liters per second?(answer)

Ex 6.2.3A ladder 13 meters long rests on horizontal ground and leansagainst a vertical wall. The foot of the ladder is pulled away fromthe wall at the rate of 0.6 m/sec. How fast is the top sliding downthe wall when the foot of the ladder is 5 m from the wall?(answer)

Ex 6.2.4A ladder 13 meters long rests on horizontal ground and leansagainst a vertical wall. The top of the ladder is being pulled up thewall at $0.1$ meters per second.How fast is the foot of the ladder approaching the wall when the foot of the ladder is 5 m from the wall?(answer)

Ex 6.2.5A rotating beacon is located 2 miles out in the water. Let $A$ be thepoint on the shore that is closest to the beacon. As the beacon rotates at10 rev/min, the beam of light sweeps down the shore once each time it revolves.Assume that the shore is straight. How fast is the point where the beamhits the shore moving at an instant when the beam is lighting up a point 2miles along the shore from the point $A$?(answer)

Ex 6.2.6A baseball diamond is a square 90 ft on a side. A player runs from firstbase to second base at 15 ft/sec. At what rate is the player's distancefrom third base decreasing when she is half way from first to second base?(answer)

Ex 6.2.7Sand is poured onto a surface at 15 cm${}^3$/sec, forming aconical pile whose base diameter is always equal to its altitude. Howfast is the altitude of the pile increasing when the pile is 3 cmhigh?(answer)

Ex 6.2.8A boat is pulled in to a dock by a rope with one end attached to the frontof the boat and the other end passing through a ring attached to the dockat a point 5 ft higher than the front of the boat. The rope is beingpulled through the ring at the rate of 0.6 ft/sec. How fast is the boatapproaching the dock when 13 ft of rope are out?(answer)

Ex 6.2.9A balloon is at a height of 50 meters, and is rising at the constant rateof 5 m/sec. A bicyclist passes beneath it, traveling in astraight line at the constant speed of 10 m/sec. How fast is the distancebetween the bicyclist and the balloon increasing 2 seconds later?(answer)

Ex 6.2.10A pyramid-shaped vat has square cross-section and stands on itstip. The dimensions at the top are 2 m $times$ 2 m, and the depth is5 m. If water is flowing into the vat at 3 m${}^3$/min, how fast isthe water level rising when the depth of water (at the deepest point)is 4 m? Note: the volume of any 'conical' shape (includingpyramids) is $(1/3)(hbox{height})(hbox{area of base})$.(answer)

Ex 6.2.11The sun is rising at the rate of $1/4$ deg/min, and appears to beclimbing into the sky perpendicular to thehorizon, as depicted in figure 6.2.5.How fast is the shadow of a 200 meter buildingshrinking at the moment when the shadow is 500 meters long? (answer)

Ex 6.2.12The sun is setting at the rate of $1/4$ deg/min, and appearsto be dropping perpendicular to the horizon, as depicted infigure 6.2.5. How fast is the shadow of a 25meter wall lengthening at the moment when the shadow is 50 meters long?(answer)

Ex 6.2.13The trough shown in figure 6.2.6is constructed by fastening together threeslabs of wood of dimensions 10 ft $times$ 1 ft, and then attaching theconstruction to a wooden wall at each end. The angle $theta$ wasoriginally $ds 30^circ$, but because of poor construction the sides arecollapsing. The trough is full of water. At what rate (in ft${}^3$/sec) is the water spilling out over the top ofthe trough if the sides have each fallen to an angle of $ds 45^circ$, and arecollapsing at the rate of $ds 1^circ$ per second?(answer)

Ex 6.2.14A woman 5 ft tall walks at the rate of 3.5 ft/sec away from a streetlightthat is 12 ft above the ground. At what rate is the tip of her shadowmoving? At what rate is her shadow lengthening?(answer)

Ex 6.2.15A man 1.8 meters tall walks at the rate of 1 meter persecond toward a streetlight that is 4 meters above the ground. Atwhat rate is the tip of his shadow moving? At what rate is his shadowshortening?(answer)

Ex 6.2.16A police helicopter is flying at 150 mph at a constant altitude of 0.5 mileabove a straight road. The pilot uses radar to determine that an oncomingcar is at a distance of exactly 1 mile from the helicopter, and that thisdistance is decreasing at 190 mph. Find the speed of the car.(answer)

Ex 6.2.17A police helicopter is flying at 200 kilometers per hour ata constant altitude of 1 km above a straight road. The pilot usesradar to determine that an oncoming car is at a distance of exactly 2kilometers from the helicopter, and that this distance is decreasing at 250kph. Find the speed of the car.(answer)

Ex 6.2.18A light shines from the top of a pole 20 m high. A ball is falling 10meters from the pole, casting a shadow on a building 30 meters away,as shown in figure 6.2.7.When the ball is 25 meters from the ground it is falling at 6 metersper second. How fast is its shadow moving?(answer)

Ex 6.2.19Do example 6.2.6 assuming that the anglebetween the two roads is 120${}^circ$ instead of 90${}^circ$ (thatis, the 'north–south' road actually goes in a somewhat northwesterlydirection from $P$). Recall the law of cosines:$ds c^2=a^2+b^2-2abcostheta$.(answer)

Ex 6.2.20Do example 6.2.6 assuming thatcar A is 300 meters north of $P$, car B is 400 meters east of $P$, bothcars are going at constant speed toward $P$, and the two cars will collide in10 seconds.(answer)

Ex 6.2.21Do example 6.2.6 assuming that8 seconds ago car A started from rest at $P$ and has been picking upspeed at the steady rate of 5 m/sec${}^2$, and 6 seconds after car Astarted car B passed $P$ moving east at constant speed 60 m/sec.(answer)

Ex 6.2.22Referring again to example 6.2.6,suppose that instead of car B an airplane is flying at speed $200$km/hr to the east of $P$ at an altitude of 2 km, as depicted infigure 6.2.8. How fast is the distance betweencar and airplane changing? (answer)

Ex 6.2.23Referring again to example 6.2.6, supposethat instead of car B an airplane is flying at speed $200$km/hr to the east of $P$ at an altitude of 2 km, and that it isgaining altitude at 10 km/hr.How fast isthe distance between car and airplane changing?(answer)

Ex 6.2.24A light shines from the top of a pole 20 m high. An object is dropped fromthe same height from a point 10 m away, so that its height at time $ds t$seconds is $ds h(t)=20-9.8t^2/2$. How fast is the object's shadowmoving on the ground one second later?(answer)

Ex 6.2.25The two blades of a pair of scissors are fastened at the point $A$ asshown in figure 6.2.9. Let$a$ denote the distance from $A$ to the tip of the blade (the point $B$).Let $beta$ denote the angle at the tip of the blade that is formed by theline $ds overline{AB}$ and the bottom edge of the blade, line$ds overline{BC}$, and let $theta$ denote the angle between$ds overline{AB}$ and the horizontal.Suppose that a piece of paper is cut in such a way that the centerof the scissors at $A$ is fixed, and the paper is also fixed. As theblades are closed (i.e., the angle $theta$ in the diagram is decreased),the distance $x$ between $A$ and $C$ increases, cutting the paper.

a. Express $x$ in terms of $a$, $theta$, and $beta$.

b. Express $dx/dt$ in terms of $a$,$theta$, $beta$, and $dtheta/dt$.

c. Suppose that the distance $a$ is 20 cm, and theangle $beta$ is $ds 5^circ$. Further suppose that $theta$ isdecreasing at 50deg/sec. At the instant when $ds theta=30^circ$, find the rate (incm/sec) at which the paper is being cut.(answer)