Discussion- Calculus

Below are two students’ discussion posts. Please provide a response for each student based on the discussion requirement below. I do not need the initial post, this is already completed, I just need the two responses.

For your response posts, you must do the following: Comment on your classmates’ analyses and their answers. Compare and contrast your problem-solving approach to how your classmates solved the problem.
Review the explanations given by your peers for their problem-solving strategies. Your comments may focus on the following:
How did they describe steps to make their explanations clear?
What additional details could they have included?
What details did they include that you may not have?
What changes would you make to your initial post?
Reply to at least two different classmates outside of your own initial post thread.

I have attached the initial post for you to use as a reference.

Student one:

The height of the cylinder is 4 inches.

A cylinder (round can) has a circular base and a circular top with vertical sides in between. Let r be the radius of the top of the can and let h be the height. The surface area of the cylinder, A, is A=2 πr2+2πrh (it’s two circles for the top and bottom plus a rolled-up rectangle for the side).

Part a: Assume that the height of your cylinder is 4 inches. Consider A as a function of r, so we can write that as A(r)=2πr2+8πr. What is the domain of A(r)? In other words, for which values of r is A(r) defined?

The domain of A(r) should be restricted to all numbers greater than 0, due to the nature of a cylinder. It cannot be a negative number in order to prevent a negative value for height and length. So the domain of A(r) in interval notation is (0, infinity).

Part b: Continue to assume that the height of your cylinder is 8inches. Write the radius r as a function of A. This is the inverse function to A(r), i.e to turn A as a function of r into. r as a function of A.

To find the inverse of A(r), A(r)=2πr^2+8πr should be written as a=2πr2+8πr in order to solve for r. From here you can subtract a in order to get 0=2πr2+8πr-a. Using the quadratic formula you can solve for r.

Quadratic Formula: x = (-b(+||-)sqrt(b^2-4ac))/2a

Once plugged in: r = (-8π(+||-)sqrt((8*π)^2-4(2π)(-a)))/2(2π)

Simplified: r = (-8π(+||-)sqrt(64*π^2+8πa))/4π

Unfortunately from here, the algebra gets a little hairy for me, and I am unable to get the correct answer in Mobius. What I did was I tried to simplify further coming up by doing this:

r = -8π/4π (+||-) sqrt((64π^2)/(16π^2) + (8πa)/(16π^2))

Simplifying to: r = -2 (+||-) sqrt(4 + (a/(2π))

Leaving my final answer to r(A) = -2 + sqrt(4 + (a/(2π)) due to the restricted domain leaving positive numbers only.

I would love it if anyone who was able to solve this could help me figure out what I am doing wrong in this step, as well as suggestions in the future for similar discussion boards. I do realize the format I am using is not intuitive to read, so I apologize and I will try to find a better way to show my work in the future.

Part c: If the surface area is 225 square inches, then what is the radius r? In other words, evaluate r(300). Round your answer to 2 decimal places.

If I use the equation I came up with, and plugin 300 for a then I believe it would look like this:

r(300) = -2 + sqrt(4 + (300/(2π))

Simplifying to r(300) = 5.19 (rounded to 2 decimal places)

The radius is 5.19 inches if the surface area is 300 square inches.

This is also incorrect, so again any suggestions to improve my understanding of the problem will be greatly appreciated! I feel like I’m either way off or I am missing one small thing. Thank you!

Student two:

Part a: Assume that the height of your cylinder is 8 inches. Consider A as a function of r, so we can write that as A(r)=2πr2+16πr. What is the domain of A(r)? In other words, for which values of r is A(r) defined? In interval notation, the values for A(r) that are defined are (0, infinity) because it does not make sense to have a negative radius or height so any value below zero would be not be defined and with a radius or height of 0 it would make the shape 2 dimensional so that would not make sense either. Therefore everything greater than 0 is defined.

Part b: Continue to assume that the height of your cylinder is 8 inches. Write the radius r as a function of A. This is the inverse function to A(r), i.e to turn A as a function of r into. r as a function of A.

To solve this, I used the quadratic formula. To make it a quadratic equation first, I have to subtract both sides by A. So A=2(pi)r^2+2(pi)RH becomes 0=2(pi)r^2+2(pi)RH-A with h=8 inches making it 0=2(pi)r^2+16(pi)r-A. This makes a=2pi; b=16pi and c=A.

Using these inputs and the quadratic formula that makes it r(A)=(-16pi+/-sqrt(((16pi)^2)-4(2pi)(-A)))/2(2pi). It doesn’t make sense to have a negative value for this as the domain is above 0 so you can rule out the “or -” and after simplifying that equation you get

r(A)=(-16pi+sqrt(256(pi^2)+8piA))/4pi.

To figure this last part out you just plug in 275 for A above and solve. so -16*pi=-50.265; 256*(pi^2)=2526.618; 8*pi*275=6911.503 and 4*pi=12.566. This makes the equation (-50.265+sqrt(2526.618+6911.503))/12.566. Then you add 2526.618+6911.503=9438.121, then you square root that which gives you an overall equation of (-50.265+97.150)/12.566 which =46.893/12.566=3.73 inches so r(275)=3.73inches meaning that for a 275 in^2 surface area you have a radius of 3.73 inches.

Requirements: 1 paragraph each

1. The domain of A(r) is ] 0; +inf [ because the radius should be bigger than 0:

If it is zero, the cylinder would be a line. And it doesn’t make sense to make a cylinder with a negative radius: there exists no such thing in real life.

Plus, A(r) = 0 has roots -6 and 0, which verifies the domain ] 0 ; +inf [

1.   r1 is always negative since its numerator is negative and its denominator is positive, while r2 is positive because it means that both the numerator and denominator are positive. The equation is, therefore:

1. We plug 125 into the equation that we found in part b) :

r(125) = = 2.375

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