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If you were the teacher

  1. Mar 23, 2006 #1

    tony873004

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    On a test, I was marked wrong on the following 2 questions.

    If you were the teacher, would you have marked me wrong?

    1. State the definition of the 1st derivative.
    My answer:
    Function that gives the slope of the tangent line for any value of x that exists.
    My score on this question: 0/2

    The answer the teacher wanted:
    [tex]
    f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}

    [/tex]
    Obviously, although he asked for the definition, he wanted the mathematical definition, and gave no points to anyone who gave a word-based definition.

    2. Suppose and object is thrown down from the top of a building at 100 ft/s. The building 1100 feet high. Its height s(t) after t seconds is given by s(t) = 1100-100t - 16t2.
    a. Find the object's average velocity on the interval [1, 3]
    What I did:
    s'(t)=-100-32t
    s'(1)=-100-32(1)=-132
    s'(3)=-100-32(3)=-196

    [tex]\frac{-132+(-136)}{2}=-164 ft/s[/tex]
    My score 1.5/2

    What he wanted:
    [tex]
    \frac{s(3)-s(1)}{3-1} =


    \frac{1100-100(3)-16(3)^2-(1100-100(1)-16(1)^2)}{3-1}) = -164 ft/s

    [/tex]
    There were certain problems on this test where he told us that we may not use certain methods (i.e. no power rule allowed on certain problems), but he did not specify a method here.
     
    Last edited: Mar 23, 2006
  2. jcsd
  3. Mar 23, 2006 #2

    Galileo

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    I would've marked them the same way.

    The first question clearly asked for the definition and the limit definition is the only rigorous one. The geometric interpretation of a derivative is different from its definition. And a function is not the same as the graph of that function. (Some functions are quite ungraphable).

    Your answer to the second question is correct, but you didn't state why you could use the method you did. The definition of the average velocity over some interval [a,b] is
    [tex]\frac{s(b)-s(a)}{b-a}[/tex]

    Only when the acceleration is constant can you say that the average velocity over the interval is the average of the velocity at the start and the velocity at the end:
    [tex]\frac{s'(a)+s'(b)}{2}[/tex]. Since you didn't state that the acceleration was constant you didn't provide a complete solution. That's why you got 1.5 points out of 2, which is fair. I would've graded it the same way.
     
    Last edited: Mar 23, 2006
  4. Mar 23, 2006 #3

    tony873004

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    But acceleration is constant as given by the question:

    s(t) = 1100-100t - 16t2

    He could have given us a longer formula that contained things such as G, MEarth, REarth, but that would have been beyond the scope of this class. The formula, as he gives it implies a constant g.

    Even so, if we were to use a non-constant g, then his method fails too.

    I had a quick 10 seconds to ask him about this after class (many others were waiting to ask questions too), and he said I was marked wrong because he asked for an average, and by finding the first derivative, I was finding an instantanous velocity. His argument made no sense to me. On Monday I get a chance to spend more time to argue my point. So I appreciate you (and anyone else who cares to participate in this thread) arguing the teacher's side for me.

    btw... I don't want to give anyone the impression that I dislike my teacher. He's one of the 5 best teachers I've had in my life, and a few other students agree. He explains things clearly, or I'd be visiting this sub-forum a lot more often.
     
  5. Mar 23, 2006 #4

    0rthodontist

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    I think the second one should get full credit though the first one shouldn't get any credit. It is the case that your method works under constant acceleration and the problem concerned constant acceleration. If he wanted to mark you wrong for not understanding average velocity, he should have done so on another problem where misunderstanding would have led to a wrong answer. As it is, he has no case that you didn't completely understand everything and neglect to mention the fact of constant acceleration simply because you took it as implicit.
     
  6. Mar 23, 2006 #5

    Tide

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    FWIW, I concur with Galileo.
     
  7. Mar 23, 2006 #6

    AKG

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    You should have gotten 0 on the first. On the second, you were lucky to get 1.5/2. You didn't prove acceleration was constant. Also, you didn't prove that when acceleration is constant, the average velocity over an interval is the arithmetic mean of the initial and final velocities.
     
  8. Mar 23, 2006 #7

    tony873004

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    I wasn't asked to prove that stuff. I was asked to find the average velocity. I found the average velocity. My answer precisely agreed with the the answers of the students who were given full credit. We were not prohibited from using any methods. Both methods need to assume a constant acceleration. If they didn't, then we would have needed to be given a MUCH more accurate value than 16 (half or 32 ft/s) in the original formula or the effects of a non-constant g would be completely drown-out by a value of g given by 2 significant digits.

    I'm curious why since Galileo's reasoning was due to a non-constant g, and the teacher's method fails with a non-constant g as well. (btw... this is not an astrophysics class).

    And I have 1 more question to put before you all... I'll put it in the next post to give me time to work on my TEX.
     
    Last edited: Mar 23, 2006
  9. Mar 23, 2006 #8

    tony873004

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    Calculate the average rate of change for the indicated value of x.
    f(x)=x2-5; x=3

    Since we were not prohibited from using the power rule, my answer was:

    f'(x)=2x
    f'(3)=2*3=6

    My score: 1.25/2

    What the teacher wanted is:

    6+h.

    My reasoning as to why he's wrong:

    6+h would be the average rate of change for [x, x+h], but not for x.
    x is a single point. It's like asking what's the batting average of a player who has 1 hit in 1 at bat. It's 1/1. Here too, with only a single point, the denominator of a fraction designed to compute an average must contain a number n=number of data points.

    So the average rate of change for a single point should be the instantanous value of that point.

    Again, thanks to everyone for playing the part of the teacher in debating against me.
     
    Last edited: Mar 23, 2006
  10. Mar 23, 2006 #9
    Also, I hope you didn't write down 196 as 136 in your paper. :wink:
     
  11. Mar 23, 2006 #10

    tony873004

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    Well at least you're paying attention :tongue2: That was a typo in the middle of the problem that doesn't work its way into the answer. And, no, I didn't make a typo on my test :).
     
  12. Mar 23, 2006 #11

    Galileo

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    Well, as unfair as it may sound afterwards, the obtained answer is not the most important thing. The reasoning behind it is. The journey is more valuable than the destination (or something like that).

    The average velocity is by definition the displacement divided by the time interval:
    [tex]\bar v = \frac{\Delta s}{\Delta t}=\frac{s(b)-s(a)}{b-a}[/tex]
    This is (by definition) general. It always holds. Constant acceleration or not.
    By constrast, if the acceleration is constant, there is another way. he velocity graph is a straight line, to find the average you can simply take the average of the end velocities. But this is not general. It doesn't holds for, say s(t)=t^3 from t in [0,1].

    Therefore, your answer is incomplete without specification that the acceleration is constant, which you could show by calculating s''(t)=-32 (=constant). So in all fairness, you got a little bit less than the full 2 points. 1.5 is not bad, right? Consider you learned from it. :wink:
     
    Last edited: Mar 23, 2006
  13. Mar 23, 2006 #12

    Curious3141

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    I'd like to add the definition for the accepted meaning of "average value of a function".

    Say you have a function f(x) defined and integrable over an interval [a,b]. The average value of that function over that interval is defined as :

    [tex]f_{avg} = \frac{1}{b-a} \int_{a}^b f(x)dx[/tex]

    That applies for *any* function. If you take the same equation and apply it to a problem of motion to calculate average velocity over the interval [t1, t2], you get :

    [tex]v_{avg} = \frac{1}{t_2-t_1} \int_{t_2}^{t_1} v(t)dt[/tex]

    Keeping in mind that v(t) = ds/dt that becomes :

    [tex]v_{avg} = \frac{1}{t_2-t_1} \int_{t_2}^{t_1} \frac{ds}{dt}dt[/tex]

    [tex]v_{avg} = \frac{1}{t_2-t_1} [s(t_2) - s(t_1)][/tex]

    or, in other words, the total displacement divided by the total time. This method made no assumptions about the acceleration, and therefore can be applied in all circumstances.

    You'll be seeing more of average value of a function in physics, notably in a.c. circuit theory.
     
    Last edited: Mar 23, 2006
  14. Mar 23, 2006 #13

    tony873004

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    agreed, except that total displacement was computed assuming constant acceleration.
     
    Last edited: Mar 23, 2006
  15. Mar 23, 2006 #14
    To get full credit on that second problem, given your method, I would have expected something like, "since the acceleration is constant, average velocity is blah blah blah." If you applied your method to any old problem, it wouldn't give you the right answer in general. Since net displacement over total time is the definition of average velocity, that method will work every time.

    If the formula given had variable acceleration, would you have attempted the same method? If I was your teacher, I might not know. Thus, without the disclaimer that the method only works in constant acceleration, I agree with your teacher and Galileo with the 1.5/2 score.

    For the third problem, it isn't possible to find the average rate of change at exactly a point. There is no change to find an average rate of change over. Because the question specified an average rate of change, not an instantaneous rate of change, you are technically incorrect. Seeing as you had about half of the answer, the score your teacher gave seems fair.
     
  16. Mar 23, 2006 #15

    Curious3141

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    It's getting a little confusing. Let's simplify things :

    1) Average velocity = total displacement/total time *always* works regardless of whether acceleration is constant or variable. THe derivation of this formula (which I gave) makes no assumptions about the nature of the velocity function v(t) or its first derivative, the acceleration a(t).

    2) Average velocity = (initial velocity plus final velocity)/2 *only* works given constant acceleration. Here an assumption is made that v(t) and a(t) fit certain conditions.

    You're not wrong in using 2) to get your answer here. However, I think to be safe, you should've explicitly mentioned that you're aware that acceleration is constant here, which is why you're using this method. See, the problem is that a naive student may think in terms of "simple" averaging - arithmetic means - of two numbers, which basically entails adding them together and dividing by two. Your teacher may have gotten the impression you're thinking this way - that you can find average velocity by just adding and dividing by two. He/she might be afraid you'll carry this same method over to problems with varying accelerations (of course, that'd be wrong), so he/she thought it would be better to "nip the habit in the bud", sort of.

    If you were very clear in your mind about what you were doing, and you wouldn't use the method you did if you were given the same problem with the equation s(t) = 2t^3 -6t^2 + 5t + 7 or s(t) = 5cos(t) + t^2 or something like that, then by all means, explain to your teacher that you know when you can use the method you used.

    But to keep things simple, it's easiest to just use the total disp/total time method. Since that method is actually so much easier to use here, you should've used that.

    In fact, if you don't mind my saying so, it seems weird that you used the method you used (taking the derivative and everything), it suggests that you actually have a misconception about the way average velocity should be worked out.
     
  17. Mar 23, 2006 #16

    matt grime

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    1. When asked for a definition give one, don't give a description of an interpretation of something. Also, the description you gave doesn't actually make sense (for any value of x that exists?)
     
  18. Mar 23, 2006 #17

    AKG

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    That's wrong. The way he computed average velocity does not depend, whatsoever, on what the acceleration is like. He uses the direct definition of average velocity: (Final position - Initial position)/(Time elapsed). All you have to do is be able to plug 3 and 1 into your formula for s(t). It doesn't matter if acceleration is 32, or e, you just need to plug in numbers and use a calculator. Your method needed to assume constant acceleration, his did not. Yes, you got the right answer, but you were lucky. His solution uses the definition of average velocity, whereas yours uses a formula that only works in special cases. Since you didn't prove that you were in this special case, why would you get full marks?

    The most important thing isn't getting the right answer. A cheater could even do that. The most important thing is knowing how to get a right answer, and knowing why a given answer is right or wrong. You didn't demonstrate that you know why your answer is right, since you didn't demonstrate that you know why you can apply the formula you used.

    EDIT: In fact, your comments above, saying that his method requires a more accurate acceleration shows that you don't fully understand what's going on. In fact, accuracy has nothing to do with anything here. 16 is not given as a measured value, where there might be room for error. It is given as an exact value.
     
    Last edited: Mar 23, 2006
  19. Mar 23, 2006 #18

    nrqed

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    I agree with your teacher for one reason: your answer just leads to the obvious question: but how is the slope of a tangent line defined then? I know you will say the answer to that one is obvious, but it is not. When providing a mathematical definition of something to be calculated, it must be something that provides an unambiguous way to do the calculation. If you define in terms of something else, you must define this something else. Of course, at some point, one had to agree on what *is* already defined and what must be defined by you but it's clear in that context that 'slope of a tangent' is not clearly defined.

    As Galileo said, your method works only for constant acceleration. You shoudl have said ''I may use this here because the acceleration is constant and I can tell this because....''.
    The problem is that your teacher can not tell if you knew that this was essential to your approach or if you thought that you could always do that. He is not psychic ;-) The problem is that others could do it the same way and never understand that their calculation worked only because a was constant. And if the don't they obviously don't deserve full credit.

    Hope this makes sense.

    Patrick
     
  20. Mar 23, 2006 #19

    tony873004

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    Thank you for all the replies so far. You're helping me anticipate what the teacher might argue when I try to convince him for a few points back.

    The general consensus so far seems to be that I'm wrong about the gravity problem because my method needs constant acceleration to be valid. I agree with that, but so does his method. Let's try a different approach. I'm going to re-word the problem with different numbers, so a non-constant gravitational field does make a significant difference. I'm also going to switch to metric units:

    Suppose an object is thrown down from the top of a building at 30.48009 m/s. The building is 4000 kilometers high. It's height s(t) after t seconds is given by s(t)=13123320-30.48009t-4.9t2. Find the object's average velocity on the interval [1, 1200]

    Accounting for a non-constant gravitational field, this object falls from r(center of Earth)=10378000 meters to r=7376000 meters (altitude, 4000km to 998km) in the 1200 second interval.

    It's average velocity is (10378000-7376000)/(1200-1)=2504 meters / second.

    Let's see how my method and how his method compare to 2504 meters per second.

    My method:
    s'(t)=(-100-9.8(1) + (-100-9.8(1200))) / 2 = -5915 meters per second

    His method:
    (13123320-30.48009*(1200)-4.9*(1200)^2 - 13123320-30.48009*(1)-4.9*(1)^2)/(1200-1)=-5915 meters per second.

    We still get the same answer, and we're both wrong. His method NEEDS to assume constant gravity too.

    I want to address some of the other points made in this thread, but I have to go to work now. Thanks again for all the replies.
     
    Last edited: Mar 23, 2006
  21. Mar 23, 2006 #20

    nrqed

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    No. This is where you are wrong (please reread AKG's post).

    The definition [itex] v_{x average} = { \Delta x \over \Delta t} [/itex] is a general definition, valid for any type of motion at all. The trick [itex] v_{x average} = {v_{x f} + v_{x i} \over 2 } [/itex] is only correct for constant acceleration.

    But you are still using an example with a constant acceleration!! . It does not suffice to say "the acceleration is not constant", you must use am equation that reflects that!! You are still using an equation for s(t) that corresponds to a constant acceleration of -9.8 m/s^2! So you are saying that you are looking at motion where the acceleration is not constant and then you start using an equation describing constant acceleration! So this is inconsistent!


    Of course, because you used the wrong equation for s(t)!!! (you used one that describes a constant acceleration!)

    Look, consider instead s(t) = 1 + 2 t^3. Now calculate the average velocity between 2 and 4 seconds using both methods. They won't agree and the correct one will be the one given by the prof's approach.
    So you were lucky to get the correct answer in the test, because if it had been a function not corresponding to constant acceleration, you would have got zero point!

    Hope this clarifies things!

    Pat
     
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