Contextual Applications of Derivatives: Background You’ll Need 3

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Contextual Applications of Derivatives: Background You’ll Need 3

Write formulas to calculate the area, perimeter, and volume of different shapes

Find the Perimeter and Area of a Rectangle

A rectangle has four sides and four right angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, [latex]L[/latex], and the adjacent side as the width, [latex]W[/latex].

A rectangle is shown. Each angle is marked with a square. The top and bottom are labeled L, the sides are labeled W.

The perimeter, [latex]P[/latex], of the rectangle is the distance around the rectangle. If you started at one corner and walked around the rectangle, you would walk [latex]L+W+L+W[/latex] units, or two lengths and two widths. The perimeter then is

[latex]\begin{array}{c}P=L+W+L+W\hfill \\ \hfill \text {or} \hfill \\ P=2L+2W\hfill \end{array}[/latex]

What about the area of a rectangle? Below is a rectangular rug. It is [latex]2[/latex] feet long by [latex]3[/latex] feet wide, and its area is [latex]6[/latex] square feet. Since [latex]A=2\cdot 3[/latex], we see that the area, [latex]A[/latex], is the length, [latex]L[/latex], times the width, [latex]W[/latex], so the area of a rectangle is [latex]A=L\cdot W[/latex].

A rectangle made up of 6 squares. The bottom is 2 squares across and marked as 2, the side is 3 squares long and marked as 3.

properties of rectangles

Rectangles have four sides and four right [latex]\left(\text{90}^ \circ\right)[/latex] angles.
The lengths of opposite sides are equal.
The perimeter, [latex]P[/latex], of a rectangle is the sum of twice the length and twice the width. See the first image.

[latex]P=2L+2W \text{ or } P = 2(L+W)[/latex]

The area, [latex]A[/latex], of a rectangle is the length times the width. The area will be expressed in square units.

[latex]A=L\cdot W[/latex]

The length of a rectangle is [latex]32[/latex] meters and the width is [latex]20[/latex] meters. Find the

Perimeter
Area

Find the Area and Perimeter of a Triangle

We now know how to find the area of a rectangle. We can use this fact to help us visualize the formula for the area of a triangle. In the rectangle below, we’ve labeled the length [latex]b[/latex] and the width [latex]h[/latex], so its area is [latex]bh[/latex].

A rectangle with the side labeled h and the bottom labeled b. The center says A equals bh.

We can divide this rectangle into two congruent triangles (see the image below). Triangles that are congruent have identical side lengths and angles, and so their areas are equal. The area of each triangle is one-half the area of the rectangle, or [latex]\Large\frac{1}{2}\normalsize bh[/latex]. This example helps us see why the formula for the area of a triangle is [latex]A=\Large\frac{1}{2}\normalsize bh[/latex].

A rectangle with a diagonal line drawn from the upper left corner to the bottom right corner. The side of the rectangle is labeled h and the bottom is labeled b. Each triangle says one-half bh. To the right of the rectangle, it says "Area of each triangle A = one-half bh".

To find the area of the triangle, you need to know its base and height. The base is the length of one side of the triangle, usually the side at the bottom. The height is the length of the line that connects the base to the opposite vertex, and makes a [latex]\text{90}^ \circ[/latex] angle with the base. The image below shows three triangles with the base and height of each marked.

Three triangles. The triangle on the left is a right triangle. The bottom is labeled b and the side is labeled h. The middle triangle is an acute triangle. The bottom is labeled b. There is a dotted line from the top vertex to the base of the triangle, forming a right angle with the base. That line is labeled h. The triangle on the right is an obtuse triangle. The bottom of the triangle is labeled b. The base has a dotted line extended out and forms a right angle with a dotted line to the top of the triangle. The vertical line is labeled h.

triangle properties

For any triangle [latex]\Delta ABC[/latex], the sum of the measures of the angles is [latex]\text{180}^ \circ[/latex].

[latex]m\angle{A}+m\angle{B}+m\angle{C}=180^\circ[/latex]

The perimeter of a triangle is the sum of the lengths of the sides.

[latex]P=a+b+c[/latex]

The area of a triangle is one-half the base, [latex]b[/latex], times the height, [latex]h[/latex].

[latex]A={\Large\frac{1}{2}}bh[/latex]

A triangle, with vertices labeled A, B, and C. The sides are labeled a, b, and c. There is a vertical dotted line from vertex B at the top of the triangle to the base of the triangle, meeting the base at a right angle. The dotted line is labeled h.

Find the area of a triangle whose base is [latex]11[/latex] inches and whose height is [latex]8[/latex] inches.

Find the Circumference and Area of Circles

The properties of circles have been studied for over [latex]2,000[/latex] years. All circles have exactly the same shape, but their sizes are affected by the length of the radius, a line segment from the center to any point on the circle. A line segment that passes through a circle’s center connecting two points on the circle is called a diameter. The diameter is twice as long as the radius. The distance around a circle is called its circumference.

A circle is shown. A dotted line running through the widest portion of the circle is labeled as a diameter. A dotted line from the center of the circle to a point on the circle is labeled as a radius. Along the edge of the circle is the circumference.

Archimedes discovered that for circles of all different sizes, dividing the circumference by the diameter always gives the same number. The value of this number is pi, symbolized by Greek letter [latex]\pi[/latex] (pronounced “pie”). We approximate [latex]\pi[/latex] with [latex]3.14[/latex] or [latex]\Large\frac{22}{7}[/latex] depending on whether the radius of the circle is given as a decimal or a fraction.

If you use the [latex]\pi[/latex] key on your calculator to do the calculations in this section, your answers will be slightly different from the answers shown. That is because the [latex]\pi[/latex] key uses more than two decimal places.

properties of circles

An image of a circle is shown. There is a line drawn through the widest part at the center of the circle with a red dot indicating the center of the circle. The line is labeled d. The two segments from the center of the circle to the outside of the circle are each labeled r.

[latex]r[/latex] is the length of the radius
[latex]d[/latex] is the length of the diameter
[latex]d=2r[/latex]
Circumference is the perimeter of a circle. The formula for circumference is [latex]C=2\pi r[/latex]
The formula for area of a circle is [latex]A=\pi {r}^{2}[/latex]

A circular sandbox has a radius of [latex]2.5[/latex] feet. Find the

Circumference of the sandbox
Area of the sandbox

Find the Volume and Surface Area of Rectangular Solids

Volume measures the space a shape occupies, while surface area describes the total area of all the surfaces of a three-dimensional object. For rectangular solids, which include cubes and rectangular prisms, these measurements are based on the object’s length, width, and height.

volume and surface area of a rectangular solid

For a rectangular solid with length [latex]L[/latex], width [latex]W[/latex], and height [latex]H[/latex]:

A rectangular solid, with sides labeled L, W, and H. Beside it is Volume: V equals LWH equals BH. Below that is Surface Area: S equals 2LH plus 2LW plus 2WH.

For a rectangular solid with length [latex]14[/latex] cm, height [latex]17[/latex] cm, and width [latex]9[/latex] cm. Find the

Volume
Surface area

Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the perimeter of a rectangle
Step 3. Name. Choose a variable to represent it.	Let [latex]P[/latex] = the perimeter
Step 4. Translate. Write the appropriate formula. Substitute.
Step 5. Solve the equation.	[latex]P=64+40[/latex] [latex]P=104[/latex]
Step 6. Check.	[latex]p\stackrel{?}{=}104[/latex] [latex]20+32+20+32\stackrel{?}{=}104[/latex] [latex]104=104\checkmark[/latex]
Step 7. Answer the question.	The perimeter of the rectangle is [latex]104[/latex] meters.

Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the area of a rectangle
Step 3. Name. Choose a variable to represent it.	Let [latex]A[/latex] = the area
Step 4. Translate. Write the appropriate formula. Substitute.
Step 5. Solve the equation.	[latex]A=640[/latex]
Step 6. Check.	[latex]A\stackrel{?}{=}640[/latex] [latex]32\cdot 20\stackrel{?}{=}640[/latex] [latex]640=640\checkmark[/latex]
Step 7. Answer the question.	The area of the rectangle is [latex]640[/latex] square meters.

Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the area of the triangle
Step 3. Name. Choose a variable to represent it.	Let [latex]A[/latex] = area of the triangle
Step 4.Translate. Write the appropriate formula. Substitute.
Step 5. Solve the equation.	[latex]A=44[/latex] square inches.
Step 6. Check.	[latex]A=\frac{1}{2}bh[/latex] [latex]44\stackrel{?}{=}\frac{1}{2}(11)8[/latex] [latex]44=44\quad\checkmark[/latex]
Step 7. Answer the question.	The area is [latex]44[/latex] square inches.

1. Circumference of the sandbox
Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the circumference of the circle
Step 3. Name. Choose a variable to represent it.	Let [latex]C[/latex] = circumference of the circle
Step 4. Translate. Write the appropriate formula Substitute	[latex]C=2\pi r[/latex] [latex]C=2\pi \left(2.5\right)[/latex]
Step 5. Solve the equation.	[latex]C\approx 2\left(3.14\right)\left(2.5\right)[/latex] [latex]C\approx 15\text{ft}[/latex]
Step 6. Check. Does this answer make sense?	Yes. If we draw a square around the circle, its sides would be [latex]5[/latex] ft (twice the radius), so its perimeter would be [latex]20[/latex] ft. This is slightly more than the circle’s circumference, [latex]15.7[/latex] ft.
Step 7. Answer the question.	The circumference of the sandbox is [latex]15.7[/latex] feet.

2. Area of the sandbox
Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the area of the circle
Step 3. Name. Choose a variable to represent it.	Let [latex]A[/latex] = the area of the circle
Step 4. Translate. Write the appropriate formula Substitute	[latex]A=\pi {r}^{2}[/latex] [latex]A=\pi{\left(2.5\right)}^{2}[/latex]
Step 5. Solve the equation.	[latex]A\approx \left(3.14\right){\left(2.5\right)}^{2}[/latex] [latex]A\approx 19.625\text{ sq. ft}[/latex]
Step 6. Check. Does this answer make sense?	Yes. If we draw a square around the circle, its sides would be [latex]5[/latex] ft, as shown in part 1. So the area of the square would be [latex]25[/latex] sq. ft. This is slightly more than the circle’s area, [latex]19.625[/latex] sq. ft.
Step 7. Answer the question.	The area of the circle is [latex]19.625[/latex] square feet.

Step 2. Identify what you are looking for.	the volume of the rectangular solid
Step 3. Name. Choose a variable to represent it.	Let [latex]V[/latex] = volume
Step 4. Translate. Write the appropriate formula. Substitute.	[latex]V=LWH[/latex] [latex]V=\mathrm{14}\cdot 9\cdot 17[/latex]
Step 5. Solve the equation.	[latex]V=2,142[/latex]
Step 6. Check We leave it to you to check your calculations.
Step 7. Answer the question.	The volume is [latex]2,142[/latex] cubic centimeters.

Step 2. Identify what you are looking for.	the surface area of the solid
Step 3. Name. Choose a variable to represent it.	Let [latex]S[/latex] = surface area
Step 4. Translate. Write the appropriate formula. Substitute.	[latex]S=2LH+2LW+2WH[/latex] [latex]S=2\left(14\cdot 17\right)+2\left(14\cdot 9\right)+2\left(9\cdot 17\right)[/latex]
Step 5. Solve the equation.	[latex]S=1,034[/latex]
Step 6. Check: Double-check with a calculator.
Step 7. Answer the question.	The surface area is [latex]1,034[/latex] square centimeters.