Oh damn... I need paper
And calculator...
Okay, (a) How much sand will the tide remove from the beach during this 6-hour period? Indicate units of measure.[I'm not writing the functions, it's early and I'm lazy, ha.]
fnInt(R(t),x,0,6)≈ 31.816 cubic yards
(b) Write an expression for Y(t), the total number of cubic yards of sand on the beach at time t.
Y(t)=S(t)-R(t)+2500
(c) Find the rate at which the total amount of sand on the beach is changing at time t=4.
What is this? rate of change... instantaneous? I think I forgot how to do this.
Wait a minute. Change of y over change of t at 4
So... Y(4)/4≈2498.091/4=624.52275 cubic yards per hour
(d) For 0≤t≤6, at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.
Find critical points S'(t)-R'(t)=0 at t=3.007 and t=.505
t=3.007, S(t)-R(t)=-2.49
t=.505, S(t)-R(t)=-.253
endpoint values
t=0, S(t)-R(t)=-2
t=6, S(t)-R(t)=2.11
So absolute minimum at t=3.007
I think I got most of them wrong... If they are, someone explain please?
Well now, off to do the rest of the calculus homework... woohoo... homework...
Sunday, April 4, 2010
Saturday, March 6, 2010
Mean Value Theorem
f'(c) = [f(b)-f(a)]/(b-a)
1. This means that if you take a function, f(x)
then the slope of the secant line between two points, [f(b)-f(a)]/(b-a)
is equal to the slope of the tangent line at the point c, f'(c)
So for example:
f(x)=x²+1 (to make things easier to see)
f'(x)=2x
[f(b)-f(a)]/(b-a) where a=-2 & b=2 ==> [4-4]/(2+2)=0
f'(c) = [f(b)-f(a)]/(b-a) ==> 2c=0 ==> c=0
Basically, it means the the secant line and the tangent line are parallel to each other.
2. This only works for continuous and differentiable functions because if the function is neither, then either the point c or the slope at c may not exist, as in the function
f(x)=|x| where a=-2 and b=2 (sorry, no graph this time). In this case, the point c would have to be at 0, but since the function is not differentiable at 0, f'(c) does not exist and the mean value theorem does not apply.
1. This means that if you take a function, f(x)
then the slope of the secant line between two points, [f(b)-f(a)]/(b-a)
is equal to the slope of the tangent line at the point c, f'(c)
So for example:
f(x)=x²+1 (to make things easier to see)
f'(x)=2x
[f(b)-f(a)]/(b-a) where a=-2 & b=2 ==> [4-4]/(2+2)=0
f'(c) = [f(b)-f(a)]/(b-a) ==> 2c=0 ==> c=0
Basically, it means the the secant line and the tangent line are parallel to each other.
2. This only works for continuous and differentiable functions because if the function is neither, then either the point c or the slope at c may not exist, as in the function
f(x)=|x| where a=-2 and b=2 (sorry, no graph this time). In this case, the point c would have to be at 0, but since the function is not differentiable at 0, f'(c) does not exist and the mean value theorem does not apply.
Wednesday, February 17, 2010
f(x) from f'(x)
Woohoo... MORE graphs... (Yes, people, that was sarcasm). Okay, ANYWAYS...
1. f(x) increasing on interval [-2, 2] because f'(x)>0
f(x) decreasing on interval (-∞, -2]U[2, ∞) because f'(x)<0
2. Extrema are at x=-2--> minimum, because f'(x) changes from negative to positive and at x=2--> maximum, because f'(x) changes from positive to negative.
3. For the sake of simplicity I'll round to the nearest whole number...
Concave up on interval (-∞, -1]U[0, 1] because f"(x), aka the slope of f'(x), is positive
Concave down on interval [-1, 0]U[1, ∞) becasue f"(x) is negative
4. I want to say that f(x) might be somthing like -x^5
since it looks somewhat like -x^3 with a bunch of crazy stuff happening in the middle...
1. f(x) increasing on interval [-2, 2] because f'(x)>0
f(x) decreasing on interval (-∞, -2]U[2, ∞) because f'(x)<0
2. Extrema are at x=-2--> minimum, because f'(x) changes from negative to positive and at x=2--> maximum, because f'(x) changes from positive to negative.
3. For the sake of simplicity I'll round to the nearest whole number...
Concave up on interval (-∞, -1]U[0, 1] because f"(x), aka the slope of f'(x), is positive
Concave down on interval [-1, 0]U[1, ∞) becasue f"(x) is negative
4. I want to say that f(x) might be somthing like -x^5
since it looks somewhat like -x^3 with a bunch of crazy stuff happening in the middle...
Friday, January 15, 2010
Mindsets
Oh wow... that was interesting...
1. First off, this seems like a load of bull to me. I don't know about the rest of you, but my "mindset" changes depending on the subject/topic,it depends on my interest in it, not on whether or I believe the brain is a muscle that can expand or not. Hell, my brain is the last thing I consider when I'm faced with something new. The first thing I think is, "Will this be interesting/useful?" If not, I tend to go towards the fixed mindset, avoid challenges, obstacles and effort. If I do think the topic is interesting or useful, I go towards the growth mindset, I take the challenges as they come and put in most of my effort --- I say most because it's hard for a lazy person to ever put in ALL their effort. In other words, I don't think there is one specific mindset that a person, or at the very least, I, have. It depends on the activity, on whether it interests you or not.
2. As said, I don't believe there is one specific mindset that controls how you learn or think all the time. Isn't this part of the reason we do well with some topics and not so well with others? Personally, I go back forth at times when learning lots of things because some are more helpful or interesting than others. The only reason I even pay attention in class is that I'm sure it will help me when it comes time to take the damned AP test. Once I deem something useless it leaves my mind for a while, until I have to review the topic again.
3. To the mind is muscle thing: My reaction was, "I already knew this." You don't take two years of biology and a year's worth of psychology without learning this. And a couple biological or psychological documentaries tend to include this fact, too...
4. It shouldn't affect me much. Considering I had heard the mind being a muscle thing before, and now, again. Oh, how I hate repetition... As for the mindset stuff, that really shouldn't affect me either considering I don't think it's entirely legitimate.
1. First off, this seems like a load of bull to me. I don't know about the rest of you, but my "mindset" changes depending on the subject/topic,it depends on my interest in it, not on whether or I believe the brain is a muscle that can expand or not. Hell, my brain is the last thing I consider when I'm faced with something new. The first thing I think is, "Will this be interesting/useful?" If not, I tend to go towards the fixed mindset, avoid challenges, obstacles and effort. If I do think the topic is interesting or useful, I go towards the growth mindset, I take the challenges as they come and put in most of my effort --- I say most because it's hard for a lazy person to ever put in ALL their effort. In other words, I don't think there is one specific mindset that a person, or at the very least, I, have. It depends on the activity, on whether it interests you or not.
2. As said, I don't believe there is one specific mindset that controls how you learn or think all the time. Isn't this part of the reason we do well with some topics and not so well with others? Personally, I go back forth at times when learning lots of things because some are more helpful or interesting than others. The only reason I even pay attention in class is that I'm sure it will help me when it comes time to take the damned AP test. Once I deem something useless it leaves my mind for a while, until I have to review the topic again.
3. To the mind is muscle thing: My reaction was, "I already knew this." You don't take two years of biology and a year's worth of psychology without learning this. And a couple biological or psychological documentaries tend to include this fact, too...
4. It shouldn't affect me much. Considering I had heard the mind being a muscle thing before, and now, again. Oh, how I hate repetition... As for the mindset stuff, that really shouldn't affect me either considering I don't think it's entirely legitimate.
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