(Note: Because I don't want to write the equations over and over again, I'll refer to them as R(t), S(t), and Y(t).)
1. R(t) is the rate at which sand is removed, so how much sand was actually removed is the integral of R(t) from t=0 to t=6. Numerically, fnInt (R(t), t, 0, 6)=31.816 cubic yards (approximately).
2. The amount of sand on the beach at time t is the initial 2500 cubic yards plus the total additional sand that is added. The total amount of sand added is the amount of sand added (the integral of S(t) from t=0 to t=x) minus the amount of sand removed (the integral of R(t) from t=0 to t=x). Mathematically, it looks like Y(t)=2500+fnInt (S(t)-R(t), t, 0, x).
3. The rate at which the total amount of sand on the beach changes is the instantaneous rate of change of Y(t) (its derivative). So, at any time time, the are is nDeriv (Y(t))=nDeriv (2500+fnInt (S(t)-R(t), t, 0, x) )=S(t)-R(t). At time t=4, the rate of change of sand is S(4)-R(4)=-1.909 cubic yards/hours.
4. From above, the rate of change of sand is Y '(t)=S(t)-R(t). As with other functions, the first thing we do is find the critical points. Found on the calculator, the point is t=5.118, as that is when Y '(t)=0, so the critical points are t=0, t=5.118, and t=6. If you look on the calculator, you can see that the sign of Y '(t) is negative until t=5.118, where it becomes positive, and remains positive when it it reaches t=6, so the only possibility for a minimum is t=5.118, as the endpoints, t=0 and t=6 would both be maximums. Therefore, the minimum value of sand is Y(5.118)=2500+fnInt (S(t)-R(t), t, 0, 5.118)=2492.369 cubic yards (approximately).
Hmm....I've looked through a couple of other people's blogs, and I'm seeing a number of answers for the last two parts. If anyone has questions on what I did, or felt I did something incorrectly, please comment and let me know.
(P.S. If anyone is wondering how I input my functions into the calculator, I made Y1=S(t), Y2=R(t), and Y3=Y1-Y2.)
Sunday, April 4, 2010
Saturday, March 13, 2010
Revision of Mean Value Theorem
Well, I'm too lazy to go back and change my first post, so I'd rather just make a new post, which contains only the revisions, not the entire post all over again. If anyone wants to see the first post for some reason, the link should be on the right side of the screen somewhere.
Why not-differentiable and discontinuous functions fail the Mean Value Theorem:
Ok, so apparently why they fail needed some revision, so let's try this again. Let us think of the function f(x)=x^2, because everyone knows how it looks. We all know it passes the Mean Value Theorem, but let's see why. There are no discontinuities, cusps, corners, or anything of the sort. As we get closer to x=0, the derivative of f gets closer to 0, and as we leave 0 towards the positive side, the derivative becomes positive. That is the general reason why differentiable functions are guarenteed a point where slope of the tangent line at x=c equals the slope of the secant line: there is a smooth transition in the sign of the derivative. If we are talking about straight line, the tangent line is the line itself, so there is no worry. Ugh...explaining with words is becoming difficult, so lets look at pictures.
Ok, again, I'm terrible with the size of picture, so I'll just explain what is happening. The graph is f(x)=x^(4/5) and the interval we are considering is [-1,1]. The secant line is y=1. There is a cusp at x=0. Because it's continuous but not differentiable, it fails the Mean Value Theorem. Why? If we examine the derivative of f, we see that it's negative at x<0>0. But if we think back to f(x)=x^2, it lacks a vital charecteristic. There is no smooth transition in the derivative from negative to zero to positive, as all curves have. In this graph, the derivative is negative, goes to a point without a derivative, and then is suddenly positive. The smooth transition just isn't there. That is why there is no guarenteed value for which the slope of the tangent line equals the slope of the secant line. I'm guessing the explanation is still a little fuzzy, so let's look at another picture.
This graph is the graph of a piecewise function, composed of y=x^2+1 from [-1,0] and y=x from (0,3], where there is a jump discontinuity at x=0 and has a secant line of y=x/4+(9/4). We know it fails the mean value theorem, but why? The exact same reason as the example above. We don't see that smooth transition in the derivative from negative to zero to positive. It goes from negative to a point where there is no derivative (due to the discontinuity), and then the derivative is thrust into being positive. The derivative doesn't naturally change but changes abruptly and suddenly. It misses values for the derivative, values the derivative can't be without the smooth transition (same as in the example above) which is why functions that aren't differentiable or continuous can't have a guarenteed point where the slope of the tangent line equals the slope of the secant line.
Wow, that was a lot of writing. That should explain everything.
(P.S. You should put more videos like that, Ms. Hwang. That contraption was extremely entertaining.)
Why not-differentiable and discontinuous functions fail the Mean Value Theorem:
Ok, so apparently why they fail needed some revision, so let's try this again. Let us think of the function f(x)=x^2, because everyone knows how it looks. We all know it passes the Mean Value Theorem, but let's see why. There are no discontinuities, cusps, corners, or anything of the sort. As we get closer to x=0, the derivative of f gets closer to 0, and as we leave 0 towards the positive side, the derivative becomes positive. That is the general reason why differentiable functions are guarenteed a point where slope of the tangent line at x=c equals the slope of the secant line: there is a smooth transition in the sign of the derivative. If we are talking about straight line, the tangent line is the line itself, so there is no worry. Ugh...explaining with words is becoming difficult, so lets look at pictures.
Ok, again, I'm terrible with the size of picture, so I'll just explain what is happening. The graph is f(x)=x^(4/5) and the interval we are considering is [-1,1]. The secant line is y=1. There is a cusp at x=0. Because it's continuous but not differentiable, it fails the Mean Value Theorem. Why? If we examine the derivative of f, we see that it's negative at x<0>0. But if we think back to f(x)=x^2, it lacks a vital charecteristic. There is no smooth transition in the derivative from negative to zero to positive, as all curves have. In this graph, the derivative is negative, goes to a point without a derivative, and then is suddenly positive. The smooth transition just isn't there. That is why there is no guarenteed value for which the slope of the tangent line equals the slope of the secant line. I'm guessing the explanation is still a little fuzzy, so let's look at another picture.
This graph is the graph of a piecewise function, composed of y=x^2+1 from [-1,0] and y=x from (0,3], where there is a jump discontinuity at x=0 and has a secant line of y=x/4+(9/4). We know it fails the mean value theorem, but why? The exact same reason as the example above. We don't see that smooth transition in the derivative from negative to zero to positive. It goes from negative to a point where there is no derivative (due to the discontinuity), and then the derivative is thrust into being positive. The derivative doesn't naturally change but changes abruptly and suddenly. It misses values for the derivative, values the derivative can't be without the smooth transition (same as in the example above) which is why functions that aren't differentiable or continuous can't have a guarenteed point where the slope of the tangent line equals the slope of the secant line.
Wow, that was a lot of writing. That should explain everything.
(P.S. You should put more videos like that, Ms. Hwang. That contraption was extremely entertaining.)
Saturday, March 6, 2010
Mean Value Theorem
Wow, so much more work for a blog post. Not used to that...
1. To understand what the mean value theorem means, you need to be able to translate it into English. f '(c) refers to the derivative at x=c and [f(b)-f(a)]/(b-a) refers to the slope of the secant line through the interval [a,b]. The mean value theorem, therefore, means that if there is a differentiable function on the interval [a,b], which implies continuity, then there is a guarenteed point at which the slope of the tangent line equals the slope of the secant line for that interval. For my example, I used the equation f(x)=4cos x.
If we consider the interval to be [0,4pi], then there are numerous points in which the slope of the tangent line equals the slope of the tangent line. The slope of the secant line is 0, and so all max and min in the interval have the same slope of 0 (the calculations are easy enough, so I leave them all to you).
2. I'm not 100% sure why it only works for differentiable functions, but my reasoning is this. If the function is both continuous and differentiable, then you can get any number as a derivative, either large or small (assuming the function is a curve). If the function is a line, then there's only one possible derivative, so by all points would have the same slope as the secant line because the function is the secant line. If the function is either not differentiable (cusp, corner, discontinuity, vertical tangent) or discontinuous, it misses that crucial charecteristic that curves have, it being that the derivative can be wither large or small. Yes, my explanation isn't the best, so lets look at a case of each and see why they fail.
This is continuous but not differentiable: f(x)=abs(x). If we pretend the interval is [-5,5], then the slope of the secant line is 0. However, there are only two possibilities for derivatives in the actual graph: -1 and 1. Relating back to what I said before, it doesn't have the charecteristic that curves have that the derivative can be large or small.
The graph is f(x)=abs(x)/x (yes, I know the graph looks strange, but it's the calculator). If we take the interval to be [-1,1], on which the function isn't continuous, then the slope of the secant line is 1. However, the only option for the derivative is 0. Again, the graph lacks the charecteristic of all curves, in that the derivative can be anything, large or small.
(Note: Sorry if the pictures aren't clear, I somewhat experimented on their size.)
1. To understand what the mean value theorem means, you need to be able to translate it into English. f '(c) refers to the derivative at x=c and [f(b)-f(a)]/(b-a) refers to the slope of the secant line through the interval [a,b]. The mean value theorem, therefore, means that if there is a differentiable function on the interval [a,b], which implies continuity, then there is a guarenteed point at which the slope of the tangent line equals the slope of the secant line for that interval. For my example, I used the equation f(x)=4cos x.
If we consider the interval to be [0,4pi], then there are numerous points in which the slope of the tangent line equals the slope of the tangent line. The slope of the secant line is 0, and so all max and min in the interval have the same slope of 0 (the calculations are easy enough, so I leave them all to you).
2. I'm not 100% sure why it only works for differentiable functions, but my reasoning is this. If the function is both continuous and differentiable, then you can get any number as a derivative, either large or small (assuming the function is a curve). If the function is a line, then there's only one possible derivative, so by all points would have the same slope as the secant line because the function is the secant line. If the function is either not differentiable (cusp, corner, discontinuity, vertical tangent) or discontinuous, it misses that crucial charecteristic that curves have, it being that the derivative can be wither large or small. Yes, my explanation isn't the best, so lets look at a case of each and see why they fail.
This is continuous but not differentiable: f(x)=abs(x). If we pretend the interval is [-5,5], then the slope of the secant line is 0. However, there are only two possibilities for derivatives in the actual graph: -1 and 1. Relating back to what I said before, it doesn't have the charecteristic that curves have that the derivative can be large or small.
The graph is f(x)=abs(x)/x (yes, I know the graph looks strange, but it's the calculator). If we take the interval to be [-1,1], on which the function isn't continuous, then the slope of the secant line is 1. However, the only option for the derivative is 0. Again, the graph lacks the charecteristic of all curves, in that the derivative can be anything, large or small.
(Note: Sorry if the pictures aren't clear, I somewhat experimented on their size.)
Saturday, February 13, 2010
The function f(x) from the graph f '(x)
1. The function f(x) is increasing at (-2,0) and (0,2) and decreasing at (negative infinity,-2) and (2, infinity). A function is increasing when f '(x)>0 and decreasing when f '(x)<0. Because the given graph is that of f '(x), we only have to check the output of the graph to find out when f '(x) is positive and negative. It can be told from looking at the graph that f '(x) is positive at (-2,0) and (0,2), meaning the function f(x) is increasing at those intervals. Similarly, it can be told that f '(x) is negative at (negative infinity, -2) and (2, infinity), so the function f(x) is decreasing at those intervals.
2. There is a local minimum at x=-2 and a local maximum at x=2. Extrema can only occur at critical points, or when f '(x)=0 or is undefined. In this case, from looking at the graph, those points would be at x=-2, x=0, and x=2. For there to be an extrema, the sign of f '(x) must change. Because it fails to do so at x=0, it isn't an extrema. At x=-2, the sign of f '(x) changes from negative to positive, so it must be a local minimum. At x=2, the sign of f '(x) changes from positive to negative, so it must be a local maximum.
3. The function f(x) is concave up at (negative infinity,-1.5) and (0, 1.5) and is concave down at (-1.5, 0) and (1.5, infinity). (Note: the values -1.5 and 1.5 are approximations, not the exact values). A function is concave up when f "(x)>0 and concave down when f "(x)<0.>0) and is concave down when the graph of f '(x) is decreasing (f "(x)<0). Just by looking at the graph, f '(x) is increasing at (negative infinity, -1.5) and (0, 1.5), so that is when the graph of f(x) is concave up. In the same manner, f '(x) is decreasing at (-1.5, 0) and (1.5, infinity), so that is when the graph of f(x) is concave down.
4. I would say f(x) is a fifth power polynomial equation. Normally, the graph of a derivative function is one degree less that the original function. The derivative of looks as if it is fourth power polynomial equation, so one can guess f(x) is a fifth power polynomial equation.
2. There is a local minimum at x=-2 and a local maximum at x=2. Extrema can only occur at critical points, or when f '(x)=0 or is undefined. In this case, from looking at the graph, those points would be at x=-2, x=0, and x=2. For there to be an extrema, the sign of f '(x) must change. Because it fails to do so at x=0, it isn't an extrema. At x=-2, the sign of f '(x) changes from negative to positive, so it must be a local minimum. At x=2, the sign of f '(x) changes from positive to negative, so it must be a local maximum.
3. The function f(x) is concave up at (negative infinity,-1.5) and (0, 1.5) and is concave down at (-1.5, 0) and (1.5, infinity). (Note: the values -1.5 and 1.5 are approximations, not the exact values). A function is concave up when f "(x)>0 and concave down when f "(x)<0.>0) and is concave down when the graph of f '(x) is decreasing (f "(x)<0). Just by looking at the graph, f '(x) is increasing at (negative infinity, -1.5) and (0, 1.5), so that is when the graph of f(x) is concave up. In the same manner, f '(x) is decreasing at (-1.5, 0) and (1.5, infinity), so that is when the graph of f(x) is concave down.
4. I would say f(x) is a fifth power polynomial equation. Normally, the graph of a derivative function is one degree less that the original function. The derivative of looks as if it is fourth power polynomial equation, so one can guess f(x) is a fifth power polynomial equation.
Wednesday, January 13, 2010
Mindsets
Wow, I didn't like this post...at all. With that being said...
1. When it comes to intelligence, I would say I'm the growth mindset...somewhat. I'm not quite sure why, but I have two conflicting answers. Based on the survey questions, I'm the growth mindset. According to the article, however, I'm of the fixed mindset. I do believe people can change their intelligence, but I try to avoid rediculously hard challenges and don't put in as much effort into things as I can. If the challenge seems doable, I'll take it. If I foresee disaster, I probably won't even attempt it (like I did with AP World History, which I dropped in a week). For those wondering how I've made it this far in math without 100% effort on my part, it's because I understand it the first time something is explained to me (usually), so I never really have problems. The times I do have problems, someone else usually asks them because they are also wondering it, so I myself never have to do the asking. I wonder if it's just that I'm lazy and laziness is what preventing me from putting in 100% effort? Well, moving on...
2. To be honest, I wouldn't know how my mindset has helped or hurt me in math. The reason being is that because I usually understand it the first time, I haven't really had problems. I think that for me to answer this question, I would need a math class that loses me completely to see what I'd do.
3. It doesn't really affect me. If it ever comes down to me actually having to train myself to study for my classes (if it has any relevance, I don't study for my classes), then I would be extremely shocked. Until I actually have to train my brain to do something, that really has no impact on me.
4. This will affect me when I come across a math class (any class for that matter) that I actually have to put in 100% effort. When such a time comes, I'll remember this article, and hopefully I'll give it everything I can. Until then, this topic will be sitting in the back of my mind.
1. When it comes to intelligence, I would say I'm the growth mindset...somewhat. I'm not quite sure why, but I have two conflicting answers. Based on the survey questions, I'm the growth mindset. According to the article, however, I'm of the fixed mindset. I do believe people can change their intelligence, but I try to avoid rediculously hard challenges and don't put in as much effort into things as I can. If the challenge seems doable, I'll take it. If I foresee disaster, I probably won't even attempt it (like I did with AP World History, which I dropped in a week). For those wondering how I've made it this far in math without 100% effort on my part, it's because I understand it the first time something is explained to me (usually), so I never really have problems. The times I do have problems, someone else usually asks them because they are also wondering it, so I myself never have to do the asking. I wonder if it's just that I'm lazy and laziness is what preventing me from putting in 100% effort? Well, moving on...
2. To be honest, I wouldn't know how my mindset has helped or hurt me in math. The reason being is that because I usually understand it the first time, I haven't really had problems. I think that for me to answer this question, I would need a math class that loses me completely to see what I'd do.
3. It doesn't really affect me. If it ever comes down to me actually having to train myself to study for my classes (if it has any relevance, I don't study for my classes), then I would be extremely shocked. Until I actually have to train my brain to do something, that really has no impact on me.
4. This will affect me when I come across a math class (any class for that matter) that I actually have to put in 100% effort. When such a time comes, I'll remember this article, and hopefully I'll give it everything I can. Until then, this topic will be sitting in the back of my mind.
Friday, December 18, 2009
Algebra vs. Calculus
1. When you're finding the limit of a function at x=c, you're finding the value that f(x) approaches as x approaches some constant. When plugging in x=c into f(x), you're finding what the actual value of f(c) is, even if f(x) doesn't necessarily approach f(c). The two cases are the same when f(x) is continuous at x=c, as given by the definition of continuity at a point lim f(x) x->c=f(c).
2. The similarities between finding the derivative and the slope of a line is that you essentially use the same formula (m=change in y/change in x), the only slight difference being when finding the derivative, you bring the two points infinitely closer to each other so that they become the same point. Mathematically, for the derivative, you use the formula lim (change in y/change in x) h->0, h being the distance between the two points. The difference is that while they are both the slope of a line, one is the slope of a specific line while the other is the slope of any line. When finding the derivative, you are finding the slope of a line, but you are finding the slope of a tangent line to a point on a curve. When saying you are finding the slope of a line, it generally refers to any ordinary line. In other words, when saying you're finding the slope of a line, it does not tell us any of the line's special properties, if it has any.
2. The similarities between finding the derivative and the slope of a line is that you essentially use the same formula (m=change in y/change in x), the only slight difference being when finding the derivative, you bring the two points infinitely closer to each other so that they become the same point. Mathematically, for the derivative, you use the formula lim (change in y/change in x) h->0, h being the distance between the two points. The difference is that while they are both the slope of a line, one is the slope of a specific line while the other is the slope of any line. When finding the derivative, you are finding the slope of a line, but you are finding the slope of a tangent line to a point on a curve. When saying you are finding the slope of a line, it generally refers to any ordinary line. In other words, when saying you're finding the slope of a line, it does not tell us any of the line's special properties, if it has any.
Wednesday, December 9, 2009
Limits
Eh, the last post was better. Anyway, for the most part, I understand limits very well. There's just one or two things that elude me:
1. The first would be the limit of the basic trig functions (sin x, cos x, tan x, csc x, sec x, cot x) as x-> positive or negative infinity. The problem is that unlike most functions, who reach a specific number or positive or negative infinity, the trig functions don't. They simply alternate between [-1,1] (sin x and cos x), (negative infinity,-1] and [1, infinity), (csc x and sec x), or (negative infinity, infinity) (tan x and cot x). Does that mean that the limit simply does not exist?
2. #13 and #14 on page 92 also confuse me. I know what they're asking and how to generally find what they're asking for, but when I try to find it algebraically, I get something that can't be simplified. Because they're absolute value problems, you can figure out the slope from the graph, but how would you find it algebraically?They are:
Find the slope of the curve at the indicated point:
13.f(x)=absolute value (x) at: a)x=2 b)x=-3
14.f(x)=absolute value (x-2) at x=1
Hm....yea, that's about it, I don't have a third thing: that's it. If anyone has answers to these questions, I would greatly appreciate it.
1. The first would be the limit of the basic trig functions (sin x, cos x, tan x, csc x, sec x, cot x) as x-> positive or negative infinity. The problem is that unlike most functions, who reach a specific number or positive or negative infinity, the trig functions don't. They simply alternate between [-1,1] (sin x and cos x), (negative infinity,-1] and [1, infinity), (csc x and sec x), or (negative infinity, infinity) (tan x and cot x). Does that mean that the limit simply does not exist?
2. #13 and #14 on page 92 also confuse me. I know what they're asking and how to generally find what they're asking for, but when I try to find it algebraically, I get something that can't be simplified. Because they're absolute value problems, you can figure out the slope from the graph, but how would you find it algebraically?They are:
Find the slope of the curve at the indicated point:
13.f(x)=absolute value (x) at: a)x=2 b)x=-3
14.f(x)=absolute value (x-2) at x=1
Hm....yea, that's about it, I don't have a third thing: that's it. If anyone has answers to these questions, I would greatly appreciate it.
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