Applications of Functions

Problem 1: A right triangle has one side x and a hypotenuse of 10 meters. Find the area of the triangle as a function of x.

Solution to Problem 1:

If the sides of a right triangle are x and y, the area A of the triangle is given by

A = ( 1 / 2) x * y

We now need to express y in terms of x using the hypotenuse, side x and pythagora's theorem

10 ² = x ² + y ²

y = sqrt [100 - x ² ]

Substitute y by its expression in the area formula to obtain

A(x) = ( 1 / 2) x sqrt [100 - x ² ]

Problem 2: A rectangle has an area equal to 100 cm² and a width x. Find the perimeter as a function of x.

Solution to Problem 2:

If x and y are the dimensions of the rectangle, using the formula of the area we obtain

100 = x * y

The perimetr P is given by

P = 2(x + y)

Solve the equation 100 = x * y for y and substitute y in the formula for the perimeter

P(x) = 2(x + 100 / x)

Problem 3: Find the area of a square as a function of its perimeter x.

Solution to Problem 3:

The area of a square of side L is given by

A = L ²

The perimetr x of a square with side L is given by

x = 4 L

Solve the above for L and substitute in the area formula A above

A(x) = (x/4) ² = x ² / 16

Problem 4: A right circular cylinder has a radius r and a height equal to twice r. Find the volume of the cylinder as a function of r.

Solution to Problem 4:

The volume V of a right circular cylinder is given by

V = (area of base of cylinder) * (height of cylinder)

= Pi * r ² * (2 r)

= 2 Pi r ³

Problem 5: Express the length L of the chord of a circle, with given radius r = 10 cm , as a function of the arc length s.(see figure below).

Solution to Problem 5:

Using half the angle a, we can write

sin(a / 2) = (L / 2) / r

Substitute r by 10 and solve for L

L = 20 sin(a / 2)

The relationship between arc length s and central angle a is

s = r a = 10 a

Solve for a

a = s / 10

Substitute a by s / 10 in L = 20 sin(a / 2) to obtain

L = 20 sin ( (s / 10) / 2 )

= 20 sin ( s / 20)

Problem 6: Express the distance d = d1+ d2, in the figure below, as a function of x.

Solution to Problem 6:

d1 is the length of the hypotenuse of a right triangle of sides x and 3, hence

d1 = sqrt[ 3 ² + x ² ]

d2 is the length of the hypotenuse of a right triangle of sides 7 - x and 5, hence

d2 = sqrt[ 5 ² + (7 - x) ² ]

d = d1 + d2 is given by

d = sqrt[ 9 + x ² ] + sqrt[ 25 + (7 - x) ² ]

Exercises

1. Express the area A of a disk in terms of its circumference C.

2. The width of a rectangle is w. Express the area A of this rectangle in terms of its perimeter P and width w.

Solutions to above exercises

1. A = C ² / (4 Pi)

2. A = (1/2) w (P - 2w)