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# Nathan's Calculus Final

Published on: Mar 3, 2016
Published in: Technology      Education

#### Transcripts - Nathan's Calculus Final

• 1. I c e a n d D u mm i e s : Comprehensive Calculus Problems from a Noteworthy Calculus student from a Prestigious High School in a Warm Climate where the people are fun
• 2. Practical Design: The Ice Column <ul><li>A 10ft Ice Column, with an initial volume of 10 π ft 3 , has a radius decreasing at 1 / 7 the rate of its height. If the volume is decreasing at a rate of .042 ft / min , what is the radius when the height is 5 ft? Assume all rates are constant. </li></ul>Related Rates The Problem:
• 3. <ul><li>Step 1: The formula for the volume of a column is V = π r ²h </li></ul><ul><li>Find the initial radius when h=10 (height), and V=10 π (volume) </li></ul><ul><li>10 π = π r ²(10) </li></ul><ul><li>r = 1 </li></ul><ul><li>Step 2: Find the Derivative of the volume of the column. You will have to use the product rule. </li></ul><ul><li>V = π r ²h </li></ul><ul><li>DV/DT = 2 π r dr / dt + π r² dh / dt </li></ul>The Solution:
• 4. Solution Cont… <ul><li>Step 3: Substitute dr / dt (the rate the radius is changing) with 1 / 7 dh / dt (the rate the height is changing), which was given in the problem. Your new equation reads as follows: </li></ul><ul><li>DV/DT = 2 π r ( 1/7) dh / dt + π r² dh / dt </li></ul><ul><li>Step 4: Substitute the values given at the start of the problem and solve for dh / dt . </li></ul><ul><li>-.042 = 2 π 1(1/7) dh / dt (10) + π (1) ² dh / dt </li></ul><ul><li>-.042 = dh / dt (20 π /7 + π ) </li></ul><ul><li>-.042 = dh / dt (27 π /7) </li></ul><ul><li>-.0035 = dh / dt </li></ul>
• 5. Solution Cont… <ul><li>Step 5 (optional): Divide dh / dt by 7 to find dr / dt . </li></ul><ul><li>dr / dt = -.0005 </li></ul><ul><li>Step 6: Find r when h = 5. Plug in all known values in the DV/DT formula and solve for r. You will need the Quadratic Formula. </li></ul><ul><li>DV/DT = 2 π rh dr / dt + π r² dh / dt </li></ul><ul><li>-.042 = 2 π (5)(-.0005)r + π r²(-.0035) </li></ul><ul><li>-.042= -.005 π r - .0035 π r² </li></ul><ul><li>0 = .0035 π r² + .005 π r - .042 </li></ul><ul><li>Step 7: Quadratic Formula, r = </li></ul><ul><li>-.005 π ± √ ( (.005 π )² - 4(.0035 π )(-.042) )/( 2(.0035 π ) ) </li></ul><ul><li>r = 1.367, -2.795 </li></ul><ul><li>Radius cannot be negative, r = 1.367ft </li></ul>
• 6. Reflection: <ul><li>Choosing a topic for presentation was a difficult process. Initially, I sought to solve a problem from my weakest area in Calculus. I wanted finally understand a topic that had troubled me most throughout the class - related rates. For most of the class and even the culminating AP exam, figuring out and solving Related Rates problems confused me. I usually did not know where to start or how to sort through the mess of scratch-work on my papers. Thus, I tested myself with a complex problem. I decided a cylinder would check my calculating skills because its derivate is much more complex (having two parts) than that of a sphere or cone. However, I ran into certain problems. For one, creating the problem ended in headache. I was not clear on what details I needed to reveal in order for it to be solved correctly. Also the solution to the problem does not make much sense, logically. One would think that if the radius is decreasing at a constant rate proportional to the height, if the height has decreased so has the radius. My findings did not support this logic. According to the math work, the radius increased from its initial value of 1 when the height was 10, to 1.3 when the height was 5. The problem may lie in that the rates of change are not accurate or that they do not work with each other. I just do not know. </li></ul><ul><li>My second problem involved Riemann Sums. I thought that simply finding an integral would be less complex than having to sort through a table of values in order to approximate an integral. Also I wanted the person solving the problem to attempt to understand what the integral of the problem meant in terms of units. Hidden in the problem is also finding a derivate, or rather approximating a derivative, and the fundamental theorem of calculus. The problem applies to a variety of skills that are necessary for understanding basic knowledge of calculus. </li></ul>