Extremely simple calc problem.

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Homework Help Overview

The discussion revolves around a calculus problem involving the motion of a particle along a curve, defined by the position function x(t) = (1/12)t^4 - (1/3)t^3 - 4t^2. Participants are tasked with determining the time at which the particle's speed is greatest, with multiple choice options provided.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss differentiating the position function to find critical points, with some suggesting the use of the second derivative to identify maxima. There are questions about the necessity of showing work for tests and the implications of using calculators versus manual calculations.

Discussion Status

The conversation is ongoing, with various interpretations of the problem being explored. Some participants express uncertainty about the correctness of their answers and the nature of the question itself, while others provide insights into the relationship between position, velocity, and acceleration.

Contextual Notes

There is mention of the problem being part of a test rather than homework, which influences participants' approaches to solving it. Additionally, the nature of the polynomial function is discussed, particularly regarding its behavior as t increases.

Liu997
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A particle moves along a curve so that its position at any time t is given by x(t) = (1/12)t^4 - (1/3)t^3 - 4t^2. For which value of t is the speed of the particle the greatest?

It was a multiple choice question so the given answers were t either equals (-2),(-1),0,2,4

Please tell me how you did it cause I would have just took the derivative and put it into my calc to see where it is highest or take the second derivative and set it equal to zero to find critical points and all that jazz.
 
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You want to find where the derivative is a maximum. This is done by computing the critical values of the derivative by computing the second derivative and setting it equal to zero (or of course finding where it is undefined). You should not need a calculator for this. Try to work things out by hand. Calculators take away from the learning experience and the fun.
 
Why don't you show us what you would do? As a rule we don't give the answers to homework or test problems.
 
Okay I differentiated to get (1/3)t^3-t^2-8t then plugged it into my ti-83. And saw -2. I'm not going to lie about actually doing it by hand because this wasn't a homework question but one on my test and I try to prevent as much work as possible. I will check it by hand though.
 
Well I took the second derivative and it is t^2-2t - 8 which easily solves to t = 4, -2
and I still get -2. -_- ughh
 
Liu997 said:
I'm not going to lie about actually doing it by hand because this wasn't a homework question but one on my test and I try to prevent as much work as possible.
(emphasis mine)
:confused: Why on Earth would you do that? I would think that the best thing to do for tests is to show as much work as possible so that one could receive partial credit if the final answer was wrong.
 
eumyang said:
(emphasis mine)
:confused: Why on Earth would you do that? I would think that the best thing to do for tests is to show as much work as possible so that one could receive partial credit if the final answer was wrong.

well it was mostly multiple choice and there was only 1 open ended and my teacher doesn't give partial credit on anything except the open ended. I had like 30 min to do like 20 problems so I figured it would not hurt to just wing it on the calc. eh I don't know
 
Liu997 said:
Well I took the second derivative and it is t^2-2t - 8 which easily solves to t = 4, -2
and I still get -2. -_- ughh

This is the 2nd derivative of the position function, but you're trying to find the maximum value of the speed (magnitude of the velocity function), so t^2 - 2t - 8 is the first derivative of the velocity. The second derivative of the velocity will give you an idea of which of the two values (t = 4, t = -2) gives the maximum velocity.
 
well I got -2 is that the correct answer to the question?
 
  • #10
If the question is at what time is the particle furthest away from the initial position, then it is the first derivative of position correct?
 
  • #11
Liu997 said:
well I got -2 is that the correct answer to the question?
Yes, that's what I get.
 
  • #12
Liu997 said:
If the question is at what time is the particle furthest away from the initial position, then it is the first derivative of position correct?
You want to find the largest value of x(t), so you would set x'(t) = 0 to find the critical points.

On the other hand, since x(t) = (1/12)t4 - (1/3)t3 - 8t2, since this is a fourth-degree polynomial with a positive leading coefficient, there won't be a maximum value. As t gets larger, the t4 term gets larger and eventually dwarfs the lower-degree terms.
 
  • #13
Mark44 said:
Yes, that's what I get.
Well this is a bummer

[URL]https://dl-web.dropbox.com/get/photo.JPG?w=06da0233[/URL]
 
Last edited by a moderator:
  • #14
Mark44 said:
You want to find the largest value of x(t), so you would set x'(t) = 0 to find the critical points.

On the other hand, since x(t) = (1/12)t4 - (1/3)t3 - 8t2, since this is a fourth-degree polynomial with a positive leading coefficient, there won't be a maximum value. As t gets larger, the t4 term gets larger and eventually dwarfs the lower-degree terms.

So is this like a poor question?
 
  • #15
No, I don't see anything wrong with the question.
 
  • #16
Liu997 said:
Well this is a bummer

[URL]https://dl-web.dropbox.com/get/photo.JPG?w=06da0233
[/URL]
Why is that? Also, your picture isn't showing up.
 
Last edited by a moderator:

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