calci relatedrates solutions

Calculus I

Preface Here are the solutions to the practice problems for my Calculus I notes. Some solutions will have more or less detail than other solutions. The level of detail in each solution will depend up on several issues. If the section is a review section, this mostly applies to problems in the first chapter, there will probably not be as much detail to the solutions given that the problems really should be review. As the difficulty level of the problems increases less detail will go into the basics of the solution under the assumption that if youve reached the level of working the harder problems then you will probably already understand the basics fairly well and wont need all the explanation. This document was written with presentation on the web in mind. On the web most solutions are broken down into steps and many of the steps have hints. Each hint on the web is given as a popup however in this document they are listed prior to each step. Also, on the web each step can be viewed individually by clicking on links while in this document they are all showing. Also, there are liable to be some formatting parts in this document intended for help in generating the web pages that havent been removed here. These issues may make the solutions a little difficult to follow at times, but they should still be readable.

Related Rates 1. In the following assume that x and y are both functions of t. Given 2x = , 1y = and 4x = determine y for the following equation.

2 2 3 4 46 2 yy x x + = e Hint : This is just like the problems worked in the section notes. The only difference is that youve been given the equation and all the needed information and so you wont have to worry about finding that. Step 1 The first thing that we need to do here is use implicit differentiation to differentiate the equation with respect to t. 2 4 4 3 4 412 2 3 4y yy y x x x x x y + = +e e Step 2 All we need to do now is plug in the given information and solve for y .

81112 16 48 32y y y + = =

2007 Paul Dawkins 1 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

2. In the following assume that x, y and z are all functions of t. Given 4x = , 2y = , 1z = ,

9x = and 3y = determine z for the following equation. ( ) 3 2 2 21 5 3x y z y z x + = + Hint : This is just like the problems worked in the section notes. The only difference is that youve been given the equation and all the needed information and so you wont have to worry about finding that. Step 1 The first thing that we need to do here is use implicit differentiation to differentiate the equation with respect to t. ( ) 2 2 21 15 2 2 2x y x y z z y y z y z z x x + = + + Step 2 All we need to do now is plug in the given information and solve for z .

45727 12 15 12 8 72z z z + + = + + = 3. For a certain rectangle the length of one side is always three times the length of the other side.

(a) If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the longer side decreasing?

(b) At what rate is the enclosed area decreasing when the shorter side is 6 inches long and is decreasing at a rate of 2 inches/minute?

Hint : The equation needed here is a really simple equation. In fact, so simple it might be easy to miss (a) If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the longer side decreasing? Step 1 Lets call the shorter side x and the longer side y. We know that 2x = and want to find y . Now all we need is an equation that relates these two quantities and from the problem statement we know the longer side is three times shorter side and so the equation is, 3y x= Step 2


Calculus I

Next step is to simply differentiate the equation with respect to t. 3y x =

Step 3

Finally, plug in the known quantity and solve for what we want : 6y =

Hint : Once we have the equation for the area we can either simplify the equation as we did in this section or we can use the result from the previous step and the equation directly. (b) At what rate is the enclosed area decreasing when the shorter side is 6 inches long and is decreasing at a rate of 2 inches/minute? Step 1 Again well call the shorter side x and the longer side y as with the last part. We know that 6x = ,

2x = and want to find A . The equation well need is just the area formula for a rectangle : A xy= At this point we can either leave the equation as is and differentiate it or we can plug in 3y x= to simplify the equation down to a single variable then differentiate. Doing this gives, ( ) 23A x x= Step 2 Now we need to differentiate with respect to t. If we use the equation in terms of only x, which is probably the easiest to use we get, 6A x x =

If we use the equation in terms of both x and y we get, A x y x y = + Step 3 Now all we need to do is plug in the known quantities and solve for A . Using the equation in terms of only x is the easiest because we already have all the known quantities from the problem statement itself. Doing this gives,

( )( )6 6 2 72A = = Now lets use the equation in terms of x and y. We know that 6x = and 2x = from the problem statement. From part (a) we have 6y = and we also know that ( )3 6 18y = = . Using these gives,

( )( ) ( )( )6 6 2 18 72A = + =


Calculus I

So, as we can see both gives the same result but the second method is slightly more work, although not much more. 4. A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at a rate of 0.5 m2/sec at what rate is the radius decreasing when the area of the sheet is 12 m2 ? Step 1 Well call the area of the sheet A and the radius r and we know that the area of a circle is given by,

2A r= We know that 0.5A = and want to determine r when 12A = . Step 2 Next step is to simply differentiate the equation with respect to t. 2A r r =

Step 3 Now, to finish this problem off well first need to go back to the equation of the area and use the fact that we know the area at the point we are interested in and determine the radius at that time.

2 1212 1.9544r r

= = =

The rate of change of the radius is then,

( )0.5 2 1.9544 0.040717r r = = 5. A person is standing 350 feet away from a model rocket that is fired straight up into the air at a rate of 15 ft/sec. At what rate is the distance between the person and the rocket increasing (a) 20 seconds after liftoff? (b) 1 minute after liftoff? Step 1 Here is a sketch for this situation that will work for both parts so well put it here.


Calculus I

Step 2 In both parts we know that 15y = and want to determine z for each given time. Using the Pythagorean Theorem we get the following equation to relate y and z. 2 2 2 2350 122500z y y= + = + Step 3 Finally, lets differentiate this with respect to t and we can even solve it for z so the actual solution will be quick and simple to find.

2 2 y yz z y y zz

= =

We have now reached a point where the process will differ for each part. (a) At what rate is the distance between the person and the rocket increasing 20 seconds after liftoff? To finish off this problem all we need to do is determine y (from the speed of the rocket and given time) and z (reusing the Pythagorean Theorem).

( )( ) 2 215 20 300 300 350 212500 50 85 460.9772y z= = = + = = = The rate of change of the distance between the two is then,

( )( )300 15 9.76187460.9772

z = =

(b) At what rate is the distance between the person and the rocket increasing 1 minute after liftoff? This part is nearly identical to the first part with the exception that the time is now 60 seconds (and note that we MUST be in seconds because the speeds are in time of seconds). Here is the work for this problem.


Calculus I

( )( )

( )( )

2 215 60 900 900 350 932500 50 373 965.6604

900 1513.98007

965.6604

y z

z

= = = + = = =

= =

6. A plane is 750 meters in the air flying parallel to the ground at a speed of 100 m/s and is initially 2.5 kilometers away from a radar station. At what rate is the distance between the plane and the radar station changing (a) initially and (b) 30 seconds after it passes over the radar station? See the (probably bad) sketch below to help visualize the problem.

(a) At what rate is the distance between the plane and the radar station changing initially? Step 1 For this part we know that 100x = when 2500x = . In this case note that x must be negative because x will be decreasing in this part. Also note that we converted x to meters since all the other quantities are in meters. Here is a sketch for this part.

Step 2 We want to determine z in this part so using the Pythagorean Theorem we get the following equation to relate x and z. 2 2 2 2750 562500z x x= + = + Step 3


Calculus I

Finally, lets differentiate this with respect to t and we can even solve it for z so the actual solution will be quick and simple to find.

2 2 x xz z x x zz

= =

Step 4 To finish off this problem all we need to do is determine z (reusing the Pythagorean Theorem) and then plug into the equation from Step 3 above.

2 22500 750 6812500 250 109 2610.0766z = + = = = The rate of change of the distance between the two for this part is,

( )( )2500 100 95.78262610.0766

z

= =

(b) At what rate is the distance between the plane and the radar station changing 30 seconds after it passes over the radar station? Step 1 For this part we know that 100x = and it will be positive in this case because x will now be increasing as we can see in the sketch below.

Step 2 As with the previous part we want to determine z and equation well need is identical to the previous part so well just rewrite both it and its derivative here.

2 2 2 2750 562500

2 2

z x xx xz z x x zz

= + = +

= =

Step 3 To finish off this problem all we need to do is determine both x and z. For x we know the speed of the plane and the fact that it has flown for 30 seconds after passing over the radar station. So x is, ( )( )100 30 3000x = =


Calculus I

For z we just need to reuse the Pythagorean Theorem.

2 23000 750 9562500 750 17 3092.3292z = + = = = The rate of change of the distance between the two for this part is then,

( )( )3000 100 97.01433092.3292

z = =

7. Two people are at an elevator. At the same time one person starts to walk away from the elevator at a rate of 2 ft/sec and the other person starts going up in the elevator at a rate of 7 ft/sec. What rate is the distance between the two people changing 15 seconds later? Step 1 Here is a sketch for this part.

We want to determine z after 15 seconds given that 2x = , 7y = and assuming that they start at the same point. Step 2 Hopefully its clear that well need the Pythagorean Theorem to solve this problem so here is that. 2 2 2z x y= + Step 3 Finally, lets differentiate this with respect to t and we can even solve it for z so the actual solution will be quick and simple to find.

2 2 2 x x y yz z x x y y zz

+ = + =

Step 4


Calculus I

To finish off this problem all we need to do is determine all three lengths of the triangle in the sketch above. We can find x and y using their speeds and time while we can find z by reusing the Pythagorean Theorem.

( )( ) ( )( )2 2

2 15 30 7 15 105

30 105 11925 15 53 109.2016

x y

z

= = = =

= + = = =

The rate of change of the distance between the two people is then,

( )( ) ( )( )30 2 105 7 7.2801109.2016

z+

= =

8. Two people on bikes are at the same place. One of the bikers starts riding directly north at a rate of 8 m/sec. Five seconds after the first biker started riding north the second starts to ride directly east at a rate of 5 m/sec. At what rate is the distance between the two riders increasing 20 seconds after the second person started riding? Step 1 Here is a sketch of this situation.

We want to determine z after 20 seconds after the second biker starts riding east given that

5x = , 8y = and assuming that they start at the same point. Step 2 Hopefully its clear that well need the Pythagorean Theorem to solve this problem so here is that. 2 2 2z x y= + Step 3 Finally, lets differentiate this with respect to t and we can even solve it for z so the actual solution will be quick and simple to find.

2 2 2 x x y yz z x x y y zz

+ = + =

Step 4


Calculus I

To finish off this problem all we need to do is determine all three lengths of the triangle in the sketch above. We can find x and y using their speeds and time while we can find z by reusing the Pythagorean Theorem. Note that the biker riding east will be riding for 20 seconds and the biker riding north will be riding for 25 seconds (this biker started 5 seconds earlier).

( )( ) ( )( )2 2

5 20 100 8 25 200

100 200 50000 100 5 223.6068

x y

z

= = = =

= + = = =

The rate of change of the distance between the two people is then,

( )( ) ( )( )100 5 200 8 9.3915223.6068

z+

= =

9. A light is mounted on a wall 5 meters above the ground. A 2 meter tall person is initially 10 meters from the wall and is moving towards the wall at a rate of 0.5 m/sec. After 4 seconds of moving is the tip of the shadow moving (a) towards or away from the person and (b) towards or away from the wall? Step 1 Here is a sketch for this situation that will work for both parts so well put it here. Also note that we know that 0.5px = for both parts.

(a) After 4 seconds of moving is the tip of the shadow towards or away from the person? Step 2 In this case we want to determine sx when ( )10 4 0.5 8px = = (although it will turn out that we simply dont need this piece of information for this problem.). We can use the idea of similar triangles to get the following equation.


Calculus I

25

s s

p s

x xx x x

= =+

If we solve this for sx we arrive at,

( )25

2 2 25 5 3 p

p s s

p s s s

x x x

x x x x x

+ =

+ = =

This equation will work perfectly for us. Step 3 Differentiation with respect to t will give us, 23 psx x = Step 4 Finishing off this problem is very simple as all we need to do is plug in the known speed.

( )2 13 30.5sx = = Because this rate is negative we can see that the tip of the shadow is moving towards the person at a rate of 13 m/s. (b) After 4 seconds of moving is the tip of the shadow towards or away from the wall? Step 2 In this case we want to determine x and the equation is really simple. All we need is, p sx x x= + Step 3 Differentiation with respect to t will give us, p sx x x = + Step 4 Finishing off this problem is very simple as all we need to do is plug in the known speeds and note that we will need to result from the first part here. So we have 12px = from the problem

statement and 13sx = from the previous part.

( ) 51 12 3 6x = + =


Calculus I

Because this rate is negative we can see that the tip of the shadow is moving towards the wall at a rate of 56 m/s. 10. A tank of water in the shape of a cone is being filled with water at a rate of 12 m3/sec. The base radius of the tank is 26 meters and the height of the tank is 8 meters. At what rate is the depth of the water in the tank changing with the radius of the top of the water is 10 meters? Step 1 Here is a sketch of the cross section of the tank and it is not even remotely to scale as I found it easier to reuse an old image that I had lying around. I can be a little lazy sometimes.

We want to determine h when 10r = and we know that 12V = . Step 2 Well need the equation for the volume of a cone. 213V r h= This is a problem however as it has both r and h in it and it would be best to have only h since we need h . We can use similar triangles to fix this up. Based on similar triangles we get the following equation which can be solved for r.

134

268

r r hh= =

Plugging this into the volume equation gives, 316948V h=


Calculus I

Step 3 Next, lets differentiate this with respect to t. 216916V h h = Step 4 To finish off this problem all we need to do is determine the value of h for the time we are interested in. This can easily be done from the similar triangle equation and the fact that we know 10r = .

( ) 404 413 13 1310h r= = = The rate of change of the height of the water is then,

( )2169 40 316 13 2512 100h h h = = = 11. The angle of elevation is the angle formed by a horizontal line and a line joining the observers eye to an object above the horizontal line. A person is 500 feet way from the launch point of a hot air balloon. The hot air balloon is starting to come back down at a rate of 15 ft/sec. At what rate is the angle of elevation, , changing when the hot air balloon is 200 feet above the ground. See the (probably bad) sketch below to help visualize the angle of elevation if you are having trouble seeing it.

Step 1 Putting variables and known quantities on the sketch from the problem statement gives,


Calculus I

We want to determine when 200y = and we know that 15y = . Step 2 There are a variety of equations that we could use here but probably the best one that involves all of the known and needed quantities is,

( )tan500

y =

Step 3 Differentiating with respect to t gives,

( ) ( )2 2sec cos500 500y y

= =

Step 4 To finish off this problem all we need to do is determine the value of for the time in question. We can either use the original equation to do this or we could acknowledge that all we really need is ( )cos and we could do a little right triangle trig to determine that. For this problem well just use the original equation to find the value of .

( ) ( )1200 2500 5tan tan 0.38051radians = = = The rate of change of the angle of elevation is then,

( )215 cos 0.38051 0.02586500

= =


PrefaceRelated Rates

calci relatedrates solutions

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