Word problems of functions

Exercise 1

Three pounds of squid can be purchased at the market for dollars. Determine the equation and represent the function that defines the cost of squid based on weight.

Exercise 2

It has been observed that a particular plant’s growth is directly proportional to time. It measured when it arrived at the nursery and exactly one week later. If the plant continues to grow at this rate, determine the function that represents the plant’s growth and graph it.

Exercise 3

A car rental charge is dollars per day plus dollars per mile travelled. Determine the equation of the line that represents the daily cost by the number of miles travelled and graph it. If a total of miles was travelled in one day, how much is the rental company going to receive as a payment?

Exercise 4

When digging into the earth, the temperature rises according to the following linear equation:

.

t is the increase in temperature in degrees and h is the depth in meters. Calculate:

1. What the temperature will be at depth?

2.Based on this equation, at what depth would there be a temperature of ?

Exercise 5

The pollution level in the centre of a city at is (parts per million) and it grows in a linear fashion by (parts per million) every hour. If is pollution and t is the time elapsed after , determine:

1.The equation that relates y with t.

2. The pollution level at 4 o’clock in the afternoon.

Exercise 6

A faucet dripping at a constant rate fills a test tube with of water every minute. Form a table of values for time and capacity, determine the equation and represent it graphically.

Exercise 7

For the function and .

1. Determine the coefficients that satisfy the equation:

2. Write the equation and represent it graphically:

3. Indicate the intervals where the function has a positive and negative value.

Solution of exercise 1

Three pounds of squid can be purchased at the market for dollars. Determine the equation and represent the function that defines the cost of squid based on weight.

Since it is a directly proportional question hence, :

Solution of exercise 2

It has been observed that a particular plant’s growth is directly proportional to time. It measured when it arrived at the nursery and exactly one week later. If the plant continues to grow at this rate, determine the function that represents the plant’s growth and graph it.

Initial height =

Weekly growth =

Solution of exercise 3

A car rental charge is dollars per day plus dollars per mile travelled. Determine the equation of the line that represents the daily cost by the number of miles travelled and graph it. If a total of miles was travelled in one day, how much is the rental company going to receive as a payment?

dollars

Solution of exercise 4

When digging into the earth, the temperature rises according to the following linear equation:

.

t is the increase in temperature in degrees and h is the depth in meters. Calculate:

1. What the temperature will be at depth?

2.Based on this equation, at what depth would there be a temperature of ?

Solution of exercise 5

The pollution level in the centre of a city at is (parts per million) and it grows in a linear fashion by (parts per million) every hour. If is pollution and t is the time elapsed after , determine:

1.The equation that relates y with t.

2. The pollution level at 4 o’clock in the afternoon.

10 hours have elapsed between 6 in the morning to four in the afternoon.

Solution of exercise 6

A faucet dripping at a constant rate fills a test tube with of water every minute. Form a table of values for time and capacity, determine the equation and represent it graphically.

Time	Capacity
…	…

Solution of exercise 7

For the function and .

1. Determine the coefficients that satisfy the equation:

2. Write the equation and represent it graphically:

3. Indicate the intervals where the function has a positive and negative value.

Problem 1 :

The total cost of airfare on a given route is comprised of the base cost C and the fuel surcharge S in rupee. Both C and S are functions of the mileage m; C(m) = 0.4m + 50 and S(m) = 0.03m. Determine a function for the total cost of a ticket in terms of the mileage and find the airfare for flying 1600 miles.

Solution :

Total cost   =  C(m) + S(m)

  =  0.4m + 50 + 0.03m

T(m)  =  0.43 m + 50

To find the airfare for flying 1600 miles, we have to apply 1600 instead of m.

T(m)  =  0.43 (1600) + 50

  =  688 + 50

T(m)  =  738

So, the total cost of airfare for flying 1600 miles is 738.

Problem 2 :

A salesperson whose annual earnings can be represented by the function A(x) = 30, 000 + 0.04x, where x is the rupee value of the merchandise he sells. His son is also in sales and his earnings are represented by the function S(x) = 25, 000 + 0.05x. Find (A + S)(x) and determine the total family income if they each sell Rupees 1, 50, 00, 000 worth of merchandise.

Solution :

(A + S)(x)   =  A(x) + S(x)

A(x) = 30, 000 + 0.04x    ——-(1)

S(x) = 25, 000 + 0.05x    ——-(2)

(1) + (2)

   =  30, 000 + 0.04x + 25, 000 + 0.05x 

   A(x) + S(x)  =  55000 + 0.09x

Here x = 15000000

   A(x) + S(x)  =  55000 + 0.09(15000000)

  =  55000 +  1350000

Total income  =  1405000

So, the required income is 1405000.

Problem 3 :

The function for exchanging American dollars for Singapore Dollar on a given day is f(x) = 1.23x, where x represents the number of American dollars. On the same day the function for exchanging Singapore Dollar to Indian Rupee is g(y) = 50.50y, where y represents the number of Singapore dollars. Write a function which will give the exchange rate of American dollars in terms of Indian rupee.

Solution :

The function for exchanging American dollars for Singapore Dollar :

f(x) = 1.23x

S.D  =  1.23 (A.D)

Here «x» stands for American dollar and f(x) stands for Singapore dollar.

A.D  = S.D/1.23  —-(1)

Exchanging Singapore Dollar to Indian Rupee is

g(y) = 50.50y

I.R  =  50.50 (S.D)

S.D  =  I.R/50.50

Applying S.D  =  I.R/50.50 in(1), we get

A.D  = (I.R/50.50)/1.23

A.D =  I.R/62.115

I.R  =  62.115 AD

So, the exchange rate of American dollars in terms of Indian rupee is I.R  =  62.115 AD.

Problem 4 :

The owner of a small restaurant can prepare a particular meal at a cost of Rupees 100. He estimates that if the menu price of the meal is x rupees, then the number of customers who will order that meal at that price in an evening is given by the function D(x) = 200−x. Express his day revenue, total cost and profit on this meal as functions of x.

Solution :

Cost price of the meal  =  100

Selling price  =  x 

Number of customers  =  200 — x

1 day revenue  =  No of customers ⋅ x

  =  (200 — x) ⋅ x

 1 day revenue  =  200 x — x²

Total cost  =  Cost of meal ⋅ No of customers

  =  100 ⋅ (200 — x)

  = 20000 — 100x 

Profit  =  Total cost — 1 day revenue

(200 — x) ⋅ x — 100 ⋅ (200 — x) 

Problem 5 :

The formula for converting from Fahrenheit to Celsius temperatures is y = (5x/9) − (160/9). Find the inverse of this function and determine whether the inverse is also a function

Solution :

 y = (5x — 160)/9

9y  = 5x — 160

5x  =  9y + 160

x  =  (1/5) (9y + 160)

f^-1 (x)  =  (1/5) (9x + 160)

f^-1 (x)  =  (9x/5) + (160/5)

f^-1 (x)  =  (9x/5) + 32

Inverse is also a function.

Problem 6 :

A simple cipher takes a number and codes it, using the function f(x) = 3x−4. Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line y = x (by drawing the lines).

Solution :

f(x) = 3x−4

Let y = 3x — 4

3x  =  y + 4

x  =   (y + 4)/3

f^-1(x)  =  (x + 4)/3

Inverse is also a function.

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This Lesson (Word problems on Trigonometric functions) was created by by ikleyn(48006)  : View Source, Show
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Word problems on Trigonometric functions

Problem 1

Outside temperature over a day can be modeled as a sinusoidal function.  Suppose you know the high temperature
of  80  degrees occurs at  5 PM  and the average temperature for the day is  75  degrees.  Find the temperature at  9 AM.

Solution

The amplitude is 80-75 = 5 degrees.


The period is 24 hours.


We want represent the function as the sine function.

So, we will start the period at  (1/4)  of 24 hours before 5 pm.

Hence, the start for the period is 11 am.


Now the general formula is


    T(t) = 5*sin(2*pi*((t-11)/24)) + 75,


where t is the time (in hours) on the clock during the day in 24-hours military notation.



In particular, at t = 9 am the temperature is  


    T(9 am) =  =  =  =  =  


            =  =  = 75 - 2.5 = 72.5   degrees.       ANSWER

Problem 2

Outside temperature over a day can be modeled as a sinusoidal function.
Suppose you know the high temperature of  96  degrees occurs at  4  PM  and
the average temperature for the day is  85  degrees.  Find the temperature,  to the nearest degree,  at  7  AM.

Solution

Our 24-hour cycle of the daily temperature change starts at 4 pm, when the temperature is maximal.


So, our sinusoidal function is, actually, the cosine function, if to start counting time from 4 pm = 16:00, or


    T(t) =  + 


where "t" is the local astronomic time in the 24-hours "military" time scale in your watch.



At 7 am next day, we have   of the full cycle elapsed.


It corresponds to    on the unit circle, and   = - .


So, the temperature at 7 am  next day will be  


     =  = 85° - 0.707*11° = 77.2°.


ANSWER.  We can expect  77.2°, or 77° rounded, at 7 am next day.

Problem 3

Outside temperature over a day can be modelled as a sinusoidal function.
Suppose you know the high temperature for the day is  90  degrees and the low temperature of  80  degrees occurs at  6  AM.
Assuming  t  is the number of hours since midnight,  find an equation for the temperature,  D(t),  in terms of  t.

Solution

The difference between the highest and lowest temperature is 90 - 80 = 10 degrees;  

hence, the midline is 85 degrees and the amplitude is   = 5 degrees.



Next, if x is the time starting from 6 am (x= 0 when the temperature is lowest), then OBVIOUSLY, the temperature as a function of x is


    D(x) =           (1)


(we use the sign  " - "  to reflect cosinusoid about the midline, and we use the ratio x/24 to model the period of 24 hours).



Now, if t is the time after midnight, then x = t-6,  so the formula (1) takes the final form


    D(t) = .     (2)     ANSWER


Or, if you like to use angular degrees instead of radians, then the same formula takes an EQUIVALENT form


    D(t) = .        (3)     ANSWER


Notice that, from (2), the angular shift in the formula (2) is one fourth of the period;

therefore, the formula (2) can be transformed this way


    D(t) = .           (4)     ANSWER

Problem 4

A Ferris wheel is  20  meters in diameter and boarded from a platform that is  1  meters above the ground.
The six o’clock position on the Ferris wheel is level with the loading platform.  The wheel completes  1  full revolution in  6  minutes.
How many minutes of the ride are spent higher than  19  meters above the ground?

Solution

This Ferris wheel is a circle of the radius of 10 meters, and its center O is at the height of 11 meters above the ground.


Imagine horizontal line at the level of 19 meters over the ground.

It is 8 meters above the center.


This horizontal line intersect the circle at two points, A and B.

Consider the triangle with the vertices O, A and the point C, which is the middle of the segment AB.


This triangle is a right angled triangle.  Its hypotenuse is the radius OA of 10 meters long.

Its leg OC is of 19-11 = 8 meters long.


Hence,  cos(x) =  = 0.8,  where x is the angle AOC.


It gives  x = arccos(0.8) = 0.64 radians.


Thus the central angle AOB  is 2x = 2*0.64 = 1.28 radians.


So, the time the person will be higher than 19 meters, can be found from the proportion


     = 


    t =  =  = 1.23 minute,  or about 1 minute and 14 seconds.    ANSWER

Problem 5

A  Ferris wheel that is  20  meters in diameter makes a revolution in  60  seconds.
If the center is  11  m above the ground,  how long does it take a rider reach a height of  16  meters?

Solution

Our Ferris wheel has the radius of 20/2 = 10 meters and the lowest position at the height of 1 meter above the ground.


Let's consider this Ferris wheel as rotated anti-clockwise.



So, it rotates from position 6 pm (the hour hand) anti-clockwise to the position 3 pm by 90 degrees,

reaching the height of 11 meters above the ground, and then rotates for some additional angle "a"

anti-clockwise, such that the height is 16-11 = 5 meters above the level of the center, which is 11 meters above the ground.



This angle "a" satisfies equation  sin(a) =  = 0.5,  so  a = arcsin(0.5) = 30 degrees.


Thus the wheel turns/rotates the angle  90 + 30 degrees = 120 degrees, total.


The angular velocity of the wheel is 360 degrees in 60 seconds, or   = 6 degrees per second.


So, it will take   = 20 seconds for the rider to get the height of 16 meters above the ground,
starting from the lowest position.


ANSWER.  It will take 20 seconds.

Problem 6

A mass is suspended by a spring such that it hangs at rest  0.5 m above the ground.
The mass is raised 40 cm and released at time t = 0 seconds,  causing it to oscillate sinusoidal.
If the mass returns to the high position every  1.2 seconds,  derive an equation of the mass oscillations.

Solution

This problem is about deriving the formula of harmonic oscillations.


    The mid-line of the sinusoid is y= 0.5 m.

    The amplitude is 0.4 m.

    The period is 1.2 seconds.

    The time shift is such that using cosine function is preferable.


So, the equation of the mass oscillation is

    y =  + .

Problem 7

Find a possible formula for the trigonometric function whose values are in the following table.

    x  0   4   8   12  16   20   24
    y  3  -1  -5   -1   3   -1   -5

Solution

The minimum is -5;  the maximum is 3.

Hence, the midline is -1 and the amplitude is 4.


The table dictates to use the cosine function.


The period starts from 0 and ends at 16.


So, the trigonometry function is


    y = .


Here "-1"  represents the midline;  the factor "4"  represents the amplitude;  and the factor (x/16)  models the 16 units period.

Problem 8

State the equation for a trigonometry function of your choice
that has an amplitude of  3  units,  a period of      and a minimum at  (0, -8).

Solution

    y = -5 - 3*cos(3x).      ANSWER.



EXPLANATION


    The minimum value of -8 and the amplitude of 3 dictate that the midline is y = -5.


    The fact that the minimum is at x= 0 dictates to use the cosine function flipped/reflected about the midline.


    The argument of  (3x)  provides / models the period  .

Problem 9

Let  a,  b,  c   be three sides opposite to the angles  A,  B,  C respectively in a triangle ABC.
Find 2ac*sin((A + C — B)/2).

Solution

First,   =  =  = .

THEREFORE,   =  = cos(B).



Next, due to the cosine law for triangle ABC

     =  +  - .



It implies

    2ac*cos(B) = a^2 + c^2 - b^2



Now we can write the final identity

     = 2ac*cos(B) = .



ANSWER.   = .

Problem 10

If  p,  q  and  r  are the sides opposite to the angles  P,  Q  and  R  respectively in a triangle  PQR.
If    =   then determine the type of the triangle  PQR:
    (a)   Equilateral
    (b)   Acute Angle but not equilateral
    (c)   Obtuse Angle
    (d)   Right Angle

Solution

It is very well known fact from Geometry that for any triangle PQR with the sides p, q, r opposite to angle P,Q, R respectively,


the ratios  ,  ,    are all equal and are equal to the diameter of the circle circumscribed around the triangle PQR


      =  =  = d.



For the proof see, for example, lesson

    - Law of sines - the Geometric Proof 

in this site.



Then from the given part of the problem, we see that


     = . = d*d = d^2.


In other words, the square of the side "r"  is equal to the square of the diameter of circumscribed circle


              = .



It implies  r = d,  i.e.  the side "r"  is the diameter of the circumscribed circle about triangle PQR.


It means that the angle R is leaning on the diameter of the circumscribed circle,

i.e. triangle PQR is a right-angled triangle with the right angle at vertex R.


So, the given triangle is a right-angled triangle.


ANSWER.  PQR is a right-angled triangle.

My other lessons on calculating trig functions and solving trig equations in this site are

    — Calculating trigonometric functions of angles

    — Advanced problems on calculating trigonometric functions of angles

    — Evaluating trigonometric expressions

    — Solve these trigonometry problems without using a calculator

    — Finding the slope of the bisector to the angle formed by two given lines in a coordinate plane

    — Solving simple problems on trigonometric equations

    — Solving typical problems on trigonometric equations

    — Solving more complicated problems on trigonometric equations

    — Solving advanced problems on trigonometric equations

    — Challenging problems on trigonometric equations

    — Miscellaneous problems on solving trigonometric equations

    — Solving twisted trigonometric equations

    — Truly elegant solution to one trigonometric equation

    — Non-standard Trigonometry problems

    — Proving Trigonometry identities

    — Calculating the sum  1*sin(1°) + 2*sin(2°) + 3*sin(3°) + . . . + 180*sin(180°)

    — Find the height

    — Solving upper-league Trigonometry equations

    — Math OLYMPIAD level problems on Trigonometry

    — Trigonometry entertainment problems

    — OVERVIEW of lessons on calculating trig functions and solving trig equations

Use this file/link  ALGEBRA-II — YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.

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