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Applications of Calculus to the Physical World (14.1 – 14.3)

 

14.1          “Rates of change as derivatives with respect to time. The notation ,,etc

14.2          Exponential growth and decay: rate of change of population: the equation
, where k is the population growth constant.

14.3          Velocity and acceleration as time derivatives. Applications involving:         (i) the determination of the velocity and acceleration of a particle given its distance from a point as a function of time;                                                         (ii) the determination of the distance of a particle from a given point, given its acceleration or velocity as a function of time together with appropriate initial conditions.”            (syllabus)

 

For a more detailed description of the requirements for this topic, see the mathematics syllabus on the Board of Studies website click here

 

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Rates of Change

 

You are familiar with differentiation and you know that is the gradient of the graph, i.e. the rate of change of y with respect to x.

 

Consider a container being filled with water. The volume, V, of water is a function of time, t. i.e. V = f(t)

Suppose the container is left under a tap that is running at a constant rate of  5 litres/minute.

The volume of water, in litres, in the container after t minutes is given by V = 5t.

We can also say = 5 i.e. the rate of change of volume with respect to time is 5 litres/minute.

 

Now consider what happens if the tap does not flow at a constant rate.

Suppose the rate at which water is flowing into the container is given by V = 5t2 + 4t

Then  = 10t + 4

It can be seen that has different values, depending on the value of t.

The initial rate of flow (when t = 0)  is (10 x 0) + 4 = 4 litres/minute

After 5 minutes the rate of flow is (10 x 5) + 4 = 54 litres/minute

 

Example:

A vessel is being filled with water such that the volume V litres that it contains at time t minutes is given by V = 4t3 – 3t2.

(i) How much water does the vessel contain after 5 minutes?

(ii) How much water is added during the third minute?

(iii) At what rate is water flowing into the vessel after 4 minutes?

 

Answer:

(i)                  V = 4 x 53 – 3 x 52 = 500 – 75 = 425 litres

(ii)                The third minute is the duration between the end of 2 minutes and the end of 3 minutes. (Your third year of life was from your second birthday to your third birthday)

V = f(3) – f(2) = (4 x 33 – 3 x 32) – (4 x 23 – 3 x 22) = 81 – 20 = 61 litres

(iii)       Rate =  = 12t2 – 6t

            At 4 minutes rate = 12 x 42 – 6 x 4 =192 – 24 = 168 litres/minute

 

Exercise:

1.         A vessel is being filled with water such that the volume V litres that it contains at time t minutes is given by V = 5t3 – 4t2.

(i) How much water does the vessel contain after 5 minutes?

(ii) How much water is added during the third minute?

(iii)    At what rate is water flowing into the vessel after 4 minutes?

 

Answers:    (i) 525 litres            (ii) 75 litres       (iii) 208 litres/minute

 

 

Rates of Change – Rate Known

 

Example:

A tank is being filled with water at a rate in litres/minute given by  R = 12t2  - 6t. After 3 minutes the water is flowing at a rate of 90 litres/minute and the tank contains 81 litres.

(i) What is the initial volume of water in the tank?

(ii) How much water does the vessel contain after 5 minutes?

(iii) If the tank holds 2000 litres, will it be full after 8 minutes?

 

Answer:

V =  dt = (12t2  - 6t) dt   =   = 4t3 – 3t2 + c

 

When t = 3 V = 81 so 81 = 4 x 33 – 3 x 32 + c = 108 -27 + c = 81 + c  c = 0

 

So V = 4t3 – 3t2

 

(i) When t = 0               V = 4t3 – 3t2 = 0

 

(ii) When t = 5 V = 4t3 – 3t2 = 4 x 53 – 3 x 52 = 500 – 75 = 425 litres

 

(iii) When t = 8             V = 4t3 – 3t2 = 4 x 83 – 3 x 82 = 2048 – 192 = 1856 litres

 

                  Since 1856 < 2000 then tank will not be full.

 

Exercise:

1.         A tank is being filled with water at a rate in litres/minute given by 

R = 6t2  - 3t. After 2 minutes the water is flowing at a rate of 18 litres/minute and the tank contains 30 litres.

(i) What is the initial volume of water in the tank?

(ii) How much water does the vessel contain after 5 minutes?

(iii) If the tank holds 1800 litres, will it be full after 10 minutes?

 

2. The number of math’s problems per hour that a student can solve is given by

     R = 32 – 2t2 where R is the number per hour and t is the time in hours.

      (i) What is the initial rate at which the student is solving math’s problems?

      (ii) At what time does the rate reduce to zero?

      (iii) What is the range of R?

      (iv) What is the domain of t?

      (v) How many problems did the student solve in the first hour? (Nearest whole number)

      (vi) What is the maximum number of problems that a student can solve in one study session? (Nearest whole number)

 

3. A block of ice is in the shape of a cube of side 25 mm. It melts so that its volume decreases at a constant rate and the ice maintains its cubic shape. After 10 minutes the sides of the ice cube measure 20 mm.

      (i) At what rate is the volume of the ice cube decreasing?

      (ii) What is the relationship between the volume, V, and the time, t?

 

Answers: 1. (i) 20 litres             (ii) 232˝  litres             (iii) yes. V = 1870 litres

 

                  2. (i) 32/hr    (ii) 4 hours     (iii) 32>R>0     (iv) 4>t>0     (v) 31     (vi) 85

 

                  3. (i) = -152.5t     (ii) V = -76.25t +15625

 

 

 Motion - Rates of Change

 

Displacement, velocity, acceleration

Physics students will already know the following two definitions from their study of motion:

Velocity is the rate of change of displacement.

 

Acceleration is the rate of change of velocity.

In our mathematics course we have to apply calculus to these definitions to find the instantaneous velocity  and instantaneous acceleration  = .

 

Example: A particle moves so that its displacement x in metres from the origin is given by x = t3 + 4t2 + 2t + 3 where t is the time in seconds. Find the (i) displacement, (ii) velocity and (iii) acceleration when t = 3 seconds.

 

Answer:            (i)  x = t3 + 4t2 + 2t + 3    when t = 3, x = 27 + 36 + 6 + 3 = 72 m

                        (ii) v == 3t2 + 8t + 2   when t = 3, v = 27 + 24 + 2 = 53 m/s

                        (iii) a =  =  = 6t + 8     when t = 3, a = 18 + 8 = 26 m/s/s

Exercise:

1.         A particle moves so that its displacement x in metres from the origin is given by x = 2t3 - 4t2 + 3t - 6 where t is the time in seconds. Find the (i) displacement, (ii) velocity and (iii) acceleration when t = 2 seconds.

2.         A particle moves so that its displacement x in metres from the origin is given by x = 2t - 8 where t is the time in seconds. Show that the velocity is constant and find the velocity.

3.         A particle moves so that its displacement x in metres from the origin is given by x = t3 - 15t2 +75t + 30 where t is the time in seconds.

(i) Find the initial displacement, velocity and acceleration

(ii) Find the time when the particle is next at rest.

(iii) Find the displacement when the particle is next at rest.

(iv)  Find the distance travelled by the particle in the third second.

4.                  A particle starts from rest at the origin and travels initially to the right with an acceleration a m/s/s where a = 4t + 2.

(i)                  Find equations to describe the velocity and position of the particle at time t.

(ii)                Find the velocity and displacement at time t = 4 seconds.

(iii)               Will the particle ever return to the origin?

 

Answers:

1. (i) 0   (ii) 11m/s     (iii) 16m/s/s                      2. v == 2m/s for all t.

 

3. (i) x = 30m, v = 75m/s, a = -30m/s/s   (ii) 5 s     (iii) 155m     (iv) 19m

 

4. (i) v = 2t2 + 2t, x = t3 + t2, (ii) v = 40m/s, x = m   (iii) No. x is always positive.

 

Exponential Growth & Decay

 

If the rate of change of a quantity is proportional to the quantity itself then it is said to be increasing or decreasing exponentially.

If the quantity is increasing (exponential growth) then it can be expressed mathematically as N = Noekt and if it is decreasing it can be expresses as N = Noe-kt

where No is the number present initially (when t = 0), N is the number present at time t, k is a constant and t is the time.

 

Example: Exponential Growth

The population of a town is increasing at an annual rate of 6%. If the population now is 30 000, what will it be in 12 years time?

Answer:

The first step is to convert the percentage to a decimal.  6% = 0.06

N = Noekt         N = 30 000e0.06 x 12 = 30 000e0.72         N = 61633

 

Example: Exponential Decay

“Barium Cow” is a radioactive isotope of barium that is used in nuclear medicine. It has a half-life of 30 seconds i.e. the amount present will reduce by half every 30 seconds. How much of a 1 kg sample of barium cow would remain after 5 minutes?

Answer:

The first step is to find the value of k.

N = Noe-kt        0.5 = 1 e -30k     ln 0.5 = -30k ln e = - 30k         k =  

= = 0.0231

 

Now we can use the value of k to determine how much barium cow remains.

Since we used the time units of seconds to find k we convert 5 minutes to 300 seconds.

N and No can be any units as long as they are the same.

In this problem we will use grams. 1 kg = 1000 grams

N = Noe-kt        N = 1000 e-0.0231 x300 = 1000 e -6.93 = 1000 x9.78 x 10-4 = 0.978 gram

 

Exercise:

1.            The population of a town is increasing at an annual rate of 5%. If the population now is 45 000, what will it be in 8 years time?

2.            An isotope of polonium (polonium-218) has a half-life of 3 minutes. If there are 109 polonium atoms present in a radioactive sample, how many polonium atoms will there be an hour later?

3.            The population of a colony of bacteria doubles every 4 hours.

(i)                  Find the hourly growth constant, k, for this colony

(ii)                If there are a billion bacteria in the colony now, how many will there be in 6 hours time?

(iii)               How long will it take the billion (109) bacteria to multiply to a trillion (1012)

 

Answers:

1. 67 132 people          2. 954 atoms   

3. (i) 0.1733     (ii) 2.83 x 109 bacteria    (iii) 39.86 hours

 

 

Applications of Calculus-Problems

 

1.            In 1960 a particular block of land had a value of $1000. The value of the block since then has been given by the equation V = 50t2 + 200t + 1000 where V is the value in dollars and t is the number of years since 1960.

(i)                  What was the value of the block of land in the year 2000?

(ii)                At what rate was the value of the block of land increasing in the year 2000?

2.            Water is running into a tank and the rate at which the level in the tank is rising is given by R = 0.2(1 - ) where R is the rate in metres/minute and t is the time in minutes. If the tank was initially empty

(i)                  What was the depth of water in the tank after half an hour?

(ii)                When will water stop flowing into the tank?

(iii)               What is the maximum depth of water in the tank?

 

3.      The number of math’s problems per hour that a student can solve is given by

         R = 48 – 3t2 where R is the number per hour and t is the time in hours.

         (i) What is the initial rate at which the student is solving math’s problems?

         (ii) At what time does the rate reduce to zero?

         (iii) What is the range of R?

         (iv) What is the domain of t?

         (v) How many problems did the student solve in the first hour? (Nearest whole number)

         (vi) What is the maximum number of problems that a student can solve in one study session? (Nearest whole number)

 

4.      The displacement  in kilometers of a car from a given point is given by

x = 3t2 + 5t where t is the time in hours.                                      

         What is the (i) displacement, (ii) velocity and (iii) acceleration of the car after 4 hours? (iv) At what time will it exceed the speed limit of 60 km/h?

 

5.      The acceleration of an object moving in a straight line is given by a = 6 – 3t where a = acceleration in m/s/s and t is the time in seconds. Measurements begin as the object moves through zero displacement at 10 m/s when t = 0.

         (i) Determine equations to find the velocity and displacement at any time,t.

         (ii) Find the velocity and displacement when the acceleration is zero.

 

6.      The population of a country is 20 million and is expanding exponentially at the rate of 3% per year.

(i)                  What will the population of the country be after 10 years?

(ii)                How long will it take for the population to double?

 

7.            An isotope of Bismuth (Bismuth-241) has a half-life of 20 minutes. What percentage of a 2 gram sample of bismuth would remain after 10 minutes?

 

Answers:

1. (i) $89 000     (ii) $4200 per year

2. (i) 4.2m    (ii) 50 minutes     (iii) 5.0m

3. (i) 48 probs./hr (ii) 4 hours  (iii) 0 < R < 48   (iv) 0 < t < 4  (v) 47   (vi) 128

4. (i) 68 km  (ii) 29 km/h (iii) 6 km/h/h  (iv) 9 hrs 10 min.

5. (i) v = 6t – 1.5t2 + 10     x = 10t + 3t2 – 0.5t3   (ii) v = 16 m/s,    x = 28m

6. (i) 27 million  (ii) 23.1 years

7. 70.7%

 

Applications of Calculus-More Problems

 

1.      In 1980 a particular block of land had a value of $50 000. The value of the block since then has been given by the equation V = 100t2 + 500t + 50 000 where V is the value in dollars and t is the number of years since 1980.

(i)                  What will be the value of the block of land in the year 2010?

(ii)                At what rate will the value of the block of land be increasing in the year 2010?

2.      Water is running into a pool and the rate at which the level in the pool is rising is given by R = (2 - ) where R is the rate in metres/minute and t is the time in minutes. If the pool was initially empty

(i)                  What was the depth of water in the pool after half an hour?

(ii)                When will water stop flowing into the pool?

(iii)               What is the maximum depth of water in the pool?

 

3.      The number of math’s problems per hour that a student can solve is given by

         R = 50 – 2t2 where R is the number per hour and t is the time in hours.

         (i) What is the initial rate at which the student is solving math’s problems?

         (ii) At what time does the rate reduce to zero?

         (iii) What is the range of R?

         (iv) What is the domain of t?

         (v) How many problems did the student solve in the first hour? (Nearest whole number)

         (vi) What is the maximum number of problems that a student can solve in one study session? (Nearest whole number)

 

4.      The displacement  in kilometers of a car from a given point is given by

x = 2t2 + 12t where t is the time in hours.                                    

         What is the (i) displacement, (ii) velocity and (iii) acceleration of the car after 4 hours? (iv) At what time will it exceed the speed limit of 60 km/h?

 

5.      The acceleration of an object moving in a straight line is given by a = 12 – 4t where a = acceleration in m/s/s and t is the time in seconds. Measurements begin as the object moves through zero displacement at 6 m/s when t = 0.

         (i) Determine equations to find the velocity and displacement at any time, t.

         (ii) Find the velocity and displacement when the acceleration is zero.

 

6.      The population of a country is 300 million and is expanding exponentially at the rate of 4% per year.

(i)   What will the population of the country be after 12 years?

(ii)                How long will it take for the population to double?

(iii)                

7.      An isotope of Bismuth (Bismuth-210) has a half-life of 5 days. What percentage of a 10 gram sample of bismuth would remain after 30 days?

 

Answers:

1. (i) $155 000     (ii) $6500 per year

2. (i) 0.3m    (ii) 40 minutes     (iii) 0.4m

3. (i) 50 probs./hr (ii) 5 hours  (iii) 0 < R < 50   (iv) 0 < t < 5  (v) 49   (vi) 167

4. (i) 80 km  (ii) 28 km/h (iii) 4 km/h/h  (iv) 12 hours.

5. (i) v = 12t – 2t2 + 6     x = 6t + 6t2 – t3   (ii) v = 24 m/s,    x = 54m

6. (i) 485 million  (ii) 17.3 years

7. 1.56%