D1 - Related RatesD1 - Related Rates

IB Math HL, MCB4U - IB Math HL, MCB4U - SantowskiSantowski

(A) Review(A) Review If we have equations/expressions that involve two (or more) variables, If we have equations/expressions that involve two (or more) variables,

we can use implicit differentiation to take derivativeswe can use implicit differentiation to take derivatives

Recall that the meaning of dRecall that the meaning of dyy/d/dxx is the rate of change of is the rate of change of yy as we as we change change xx

Recall that a rate is a change of some quantity with respect to time. Recall that a rate is a change of some quantity with respect to time. As a derivative, we would write this as dAs a derivative, we would write this as dX /X / d dtt

So therefore dSo therefore dVV /d /dtt would mean => the rate of change of the volume would mean => the rate of change of the volume as time changesas time changes

Likewise dLikewise dAA/ d/ drr would mean => the rate of change of the area as we would mean => the rate of change of the area as we make changes in the radius make changes in the radius

So also consider:So also consider: ddVV / d / drr ddhh / d / dtt ddhh / d / drr ddrr / d / dtt

(B) Related Rates and (B) Related Rates and CirclesCircles

ex 1. A pebble is dropped into a pond and the ripples form ex 1. A pebble is dropped into a pond and the ripples form concentric circles. The radius of the outermost circle increases at concentric circles. The radius of the outermost circle increases at a constant rate of 10 cm/s. Determine the rate at which the area a constant rate of 10 cm/s. Determine the rate at which the area of the disturbed water is changing when the radius is 50 cm.of the disturbed water is changing when the radius is 50 cm.

So, as the radius changes, so does the area So, as the radius changes, so does the area dr/dt is related to dr/dt is related to dA/dt dA/dt how? how?

Recall the area formula Recall the area formula A = A = rr22

Then use implicit differentiation as we now differentiate with Then use implicit differentiation as we now differentiate with respect to time respect to time d/dt (A) = d/dt ( d/dt (A) = d/dt (rr22) = 2) = 2r r dr/dt dr/dt

So now we know the relationship between dA/dt and dr/dtSo now we know the relationship between dA/dt and dr/dt Then if we knew dr/dt, we could find dA/dtThen if we knew dr/dt, we could find dA/dt It is given that dr/dt is 10 cm/sec, then dA/dt = 2 It is given that dr/dt is 10 cm/sec, then dA/dt = 2 (50cm) (50cm)

10 cm/sec10 cm/sec dA/dt = 1000dA/dt = 1000 cm cm22/sec or 3141 square cm per second/sec or 3141 square cm per second

(C) Related Rates and (C) Related Rates and ConesCones

Ex 2. A water tank has a shape of an inverted circular cone Ex 2. A water tank has a shape of an inverted circular cone with a base radius of 2 m and a height of 4 m. If water is with a base radius of 2 m and a height of 4 m. If water is being pumped in a rate of 2 mbeing pumped in a rate of 2 m33/min, find the rate at which /min, find the rate at which the water level is rising when the water is 3 meters deep.the water level is rising when the water is 3 meters deep.

So in the context of this conical container filling, we see So in the context of this conical container filling, we see that the rate of change of the volume is related to 2 that the rate of change of the volume is related to 2 different rates different rates the rate of change of the height and the the rate of change of the height and the rate of change of the radiusrate of change of the radius

Likewise, the rate of change of the height is related to the Likewise, the rate of change of the height is related to the rate of change of the radius and the rate of change of time.rate of change of the radius and the rate of change of time.

So as we drain the conical container, several things change So as we drain the conical container, several things change V,r,h,t and how they change is related V,r,h,t and how they change is related

(C) Related Rates and (C) Related Rates and ConesCones

Having the formula V = 1/3Having the formula V = 1/3rr22h and the differentiation dV/dt = h and the differentiation dV/dt = d/dt(1/3d/dt(1/3rr22h), we can now take the derivativeh), we can now take the derivative

dV/dt = 1/3dV/dt = 1/3 d/dt [(r d/dt [(r2 2 h)]h)] dV/dt = 1/3dV/dt = 1/3 [ (2r [ (2r dr/dt dr/dt h) + (dh/dt h) + (dh/dt r r22)])]

So as we suspected initially, the rate at which the volume in a So as we suspected initially, the rate at which the volume in a cone changes is related to the rate at which the radius changes cone changes is related to the rate at which the radius changes and the rate at which the height changes.and the rate at which the height changes.

If we know these 2 rates (dr/dt and dh/dt) we can solve the If we know these 2 rates (dr/dt and dh/dt) we can solve the problemproblem

But if we do not know the 2 rates, we need some other But if we do not know the 2 rates, we need some other relationship to help us out.relationship to help us out.

In the case of a cone In the case of a cone we usually know the relationship between we usually know the relationship between the radius and the height and can express one in terms of the the radius and the height and can express one in terms of the otherother

(C) Related Rates and (C) Related Rates and ConesCones

In a right angled cone (radius is perpendicular to the In a right angled cone (radius is perpendicular to the height) the ratio of height to radius is always constantheight) the ratio of height to radius is always constant

In this case, h/r = 4/2 In this case, h/r = 4/2 so r = ½h so r = ½h So our formula V = 1/3So our formula V = 1/3rr22h becomes V = 1/3h becomes V = 1/3(½h)(½h)22h = h =

1/121/12hh33

Now we can differentiate againNow we can differentiate again dV/dt = d/dt (1/12dV/dt = d/dt (1/12hh33)) dV/dt = 1/12dV/dt = 1/12 3h 3h22 dh/dt dh/dt 2 m2 m33/min = 1/12/min = 1/12 3(3) 3(3)22 dh/dt dh/dt dh/dt = 2 dh/dt = 2 ÷ (27/12÷ (27/12)) dh/dt = 8dh/dt = 8/9 m/min/9 m/min

(D) Related Rates and (D) Related Rates and Pythagorean RelationshipsPythagorean Relationships

A ladder 10 meters long rests against a vertical wall. If the A ladder 10 meters long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 bottom of the ladder slides away from the wall at a rate of 1 m/s, how fast is the top of the ladder sliding down the wall m/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 m from the foot of the when the bottom of the ladder is 6 m from the foot of the wall?wall?

If we set up a diagram, we create a right triangle, where the If we set up a diagram, we create a right triangle, where the ladder represents the hypotenuse and then realize that the ladder represents the hypotenuse and then realize that the quantities that change with time are the distance of the foot quantities that change with time are the distance of the foot of the ladder along the floor (of the ladder along the floor (xx) and the distance from the top ) and the distance from the top of the ladder to the floor (of the ladder to the floor (yy))

So we let x represent this distance of the foot of the ladder So we let x represent this distance of the foot of the ladder and y represents the distance of the top of the ladder to the and y represents the distance of the top of the ladder to the floorfloor

So our mathematical relationship is xSo our mathematical relationship is x22 + y + y22 = 10 = 1022

(D) Related Rates and (D) Related Rates and Pythagorean RelationshipsPythagorean Relationships

So we have xSo we have x22 + y + y22 = 10 = 1022 and we simply and we simply differentiate wrt timedifferentiate wrt time

Then d/dt (xThen d/dt (x22 + y + y22) = d/dt (10) = d/dt (1022)) 2x 2x dx/dt + 2y dx/dt + 2y dy/dt = 0 dy/dt = 0 x x dx/dt = - y dx/dt = - y dy/dt dy/dt Then dy/dt = x Then dy/dt = x dx/dt dx/dt ÷ - y÷ - y So dy/dt = So dy/dt = (6 m) (6 m) (1 m/s) (1 m/s) ÷ -(8) = ÷ -(8) = dy/dt = -3/4 m/secdy/dt = -3/4 m/sec

So the top of the ladder is coming down at a rate So the top of the ladder is coming down at a rate of 0.75 m/sec when the foot of the ladder is 6 m of 0.75 m/sec when the foot of the ladder is 6 m from the wallfrom the wall

(D) Related Rates and (D) Related Rates and AnglesAngles

You walk along a straight path at a speed of 4 m/s. A search light You walk along a straight path at a speed of 4 m/s. A search light is located on the ground, a perpendicular distance of 20 m from is located on the ground, a perpendicular distance of 20 m from your path. The light stays focused on you. At what rate does the your path. The light stays focused on you. At what rate does the search light rotate when you are 15 meters from the point on the search light rotate when you are 15 meters from the point on the path closest to the search light?path closest to the search light?

So we need a relationship between the angle, the 20 meters and So we need a relationship between the angle, the 20 meters and your distance along the path your distance along the path use the primary trig ratios to set use the primary trig ratios to set this up this up the angle is that between the perpendicular (measuring the angle is that between the perpendicular (measuring 20 meters) and the path of the opposite side 20 meters) and the path of the opposite side opposite and opposite and adjacent are related by the tangent ratioadjacent are related by the tangent ratio

So tan(So tan() = x/20 or x = 20 tan() = x/20 or x = 20 tan() ) Differentiating Differentiating d/dt (x) = d/dt (20 tan( d/dt (x) = d/dt (20 tan()))) Thus dx/dt = 20 Thus dx/dt = 20 sec sec22(() ) d d/dt = 4 m/s/dt = 4 m/s Then dThen d/dt = 4 /dt = 4 ÷ 20÷ 20secsec22(())

(D) Related Rates and (D) Related Rates and AnglesAngles

Then dThen d/dt = 4 /dt = 4 ÷ 20÷ 20secsec22(())

To find secTo find sec22((), the measures in our triangle at ), the measures in our triangle at the instant in question are the 20m as the the instant in question are the 20m as the perpendicular distance, 15m as the distance from perpendicular distance, 15m as the distance from the perpendicular, and then the hypotenuse as the perpendicular, and then the hypotenuse as 25m 25m so sec so sec22((), = (25/15)), = (25/15)22 = 25/16 = 25/16

Then dThen d/dt = 4 /dt = 4 ÷ ÷ (20 (20 25 25 ÷ 16) = 0.128 rad/sec÷ 16) = 0.128 rad/sec

So the search light is rotating at 0.128 rad/secSo the search light is rotating at 0.128 rad/sec

(F) Summary(F) Summary

In this section we’ve seen four related rates In this section we’ve seen four related rates problems.  They all work in essentially the same problems.  They all work in essentially the same way.  The main difference between them was way.  The main difference between them was coming up with the relationship between the coming up with the relationship between the known and unknown quantities.  This is often the known and unknown quantities.  This is often the hardest part of the problem.hardest part of the problem.

   The best way to come up with the relationship is The best way to come up with the relationship is

to sketch a diagram that shows the situation.  to sketch a diagram that shows the situation.  This often seems like a silly step, but can make This often seems like a silly step, but can make all the difference in whether we can find the all the difference in whether we can find the relationship or not. relationship or not.

(G) Internet Links(G) Internet Links

Related Rates from Calculus I - Problems and SoluRelated Rates from Calculus I - Problems and Solutions by tions by PhengPheng Kim Kim VingVing

Tutorial for Related Rates from Stefan Tutorial for Related Rates from Stefan WanerWaner at at HofstraHofstra U U

Related Rates from Calculus on the Web from TeRelated Rates from Calculus on the Web from Temple U by Gerardo Mendoza and Dan Reichmple U by Gerardo Mendoza and Dan Reich

Calculus I (Math 2413) - Derivatives - Related RatCalculus I (Math 2413) - Derivatives - Related Rates from Paul Hawkins at Lamar Ues from Paul Hawkins at Lamar U

(I) Homework(I) Homework

IB Math HL, Stewart, 1989, Chap 3.5, IB Math HL, Stewart, 1989, Chap 3.5, p145, Q3-17 do evens or oddsp145, Q3-17 do evens or odds

Try also Chap 7.4, p326, Q8-14 to Try also Chap 7.4, p326, Q8-14 to work with some angle relationshipswork with some angle relationships

MCB4U, MCB4U,