Question

activity 2
building a cubic function from a
quadratic and linear function
in the previous activity, you built a cubic function from three linear
functions. in this activity, you will build a cubic function from a
quadratic function and a linear function.
the plant-a-seed company also makes cylindrical planters for city
sidewalks and storefronts. the cylindrical planters come in a variety
of sizes, but all have a height that is twice the radius.
1 why do you think plant-a-seed might want to manufacture different
sizes of a product but maintain a constant ratio of height to radius?
2 consider differently sized cylindrical planters.
(a) complete the table.
(b) describe how you determine the volume when you know the radius.
topic 2 > composing and

Explanation:

Step1: Calculate base area

The formula for the base area \(A\) of a cylinder (which is a circle) is \(A = \pi r^{2}\).

• When \(r = 0\), \(A=\pi\times0^{2}=0\)
• When \(r = 1\), \(A=\pi\times1^{2}=\pi\approx3.14\)
• When \(r = 2\), \(A=\pi\times2^{2}=4\pi\approx12.56\)
• When \(r = 3\), \(A=\pi\times3^{2}=9\pi\approx28.26\)
• When \(r = 4\), \(A=\pi\times4^{2}=16\pi\approx50.24\)
• When \(r = x\), \(A=\pi x^{2}\)

Step2: Calculate volume

The formula for the volume \(V\) of a cylinder is \(V=Ah\) (where \(A\) is the base area and \(h\) is the height). Since \(h = 2r\), we can also write \(V=\pi r^{2}\times(2r)=2\pi r^{3}\)

• When \(r = 0\), \(V = 2\pi\times0^{3}=0\)
• When \(r = 1\), \(V=2\pi\times1^{3}=2\pi\approx6.28\)
• When \(r = 2\), \(V=2\pi\times2^{3}=16\pi\approx50.24\)
• When \(r = 3\), \(V=2\pi\times3^{3}=54\pi\approx169.56\)
• When \(r = 4\), \(V=2\pi\times4^{3}=128\pi\approx401.92\)
• When \(V = 2000\), we solve \(2\pi r^{3}=2000\), \(r^{3}=\frac{2000}{2\pi}=\frac{1000}{\pi}\approx318.31\), \(r=\sqrt[3]{\frac{1000}{\pi}}\approx6.8\)

Step3: Describe volume - radius relationship

We know that the volume of a cylinder \(V=\pi r^{2}h\). Given \(h = 2r\), substitute \(h\) into the volume formula. So \(V=\pi r^{2}(2r)=2\pi r^{3}\). So if we know the radius \(r\) of the cylindrical planter, we can find the volume by using the formula \(V = 2\pi r^{3}\)

Answer:

Radius	Height (inches)	Base Area (square inches)	Volume (cubic inches)
1	2	\(\pi\approx3.14\)	\(2\pi\approx6.28\)
2	4	\(4\pi\approx12.56\)	\(16\pi\approx50.24\)
3	6	\(9\pi\approx28.26\)	\(54\pi\approx169.56\)
4	8	\(16\pi\approx50.24\)	\(128\pi\approx401.92\)
\(\sqrt[3]{\frac{1000}{\pi}}\approx6.8\)	\(\ 2\sqrt[3]{\frac{1000}{\pi}}\approx13.6\)	\(\pi(\sqrt[3]{\frac{1000}{\pi}})^{2}\approx145.1\)	2000
\(x\)	\(2x\)	\(\pi x^{2}\)	\(2\pi x^{3}\)

For part (b), if we know the radius \(r\) of the cylindrical planter, we use the formula \(V = 2\pi r^{3}\) to find the volume.