You are not logged in.
The real answer is 2. You are getting 1.9987 and 2.0077 because of inaccuracy with the graphing. I would answer 2.
Ok your right. it DOES look like a straight line. I kept changing the windows settings on my calculator but it still looked as a line. However, my TI-89 calculator has a function that finds the minimum value of a function on an interval, I did it from x=0 to x=4 and it says the minimum value is at x=2 and y=0.15
0.5(9.8)(0.87025) = 4.264225 ?
Nevermind. Your right!
Thanks.
Here is the problem:
"A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s, releases a sandbag at an instant when the balloon is a height h = 40.0 m above the ground View Figure . After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive."
Question:
Compute the position of the sandbag at a time 0.295 s after its release.
We are supposed to be doing this in some online math software and you input the answer and it validates it for you etc. I solved it but when I put my answer it says its incorrect... Here what i did:
y = y0 + V0yt + .5Ayt^2
y = final position (find)
y0 = initial position (40)
V0y = initial velocity (5 m/s)
t = time (0.295)
Ay = acceleration (gravity) = -9.8 m/s
So given the information in problem, I have to give the final position of the sandbox which corresponds to y.
y = 40 + (5 m/s)(0.295) + .5(-9.8 m/s)(0.295)^2
y = 40 +1.475 - 0.5(9.8)(0.87025)
y = 41.475 - 4.264225
y = 37.210775
y ~ 37.21
However, the online thing keeps saying its wrong? Could anyone help me? I have to turn this in by 12.
Im not sure what to tell you. As I said before, can you post the function so I can graph it myself and see?
1. For g(x) to be continuous everywhere the limit of the function from the left and from the right as x approaches c have to be the same as the function value at c. In this case c being 2. That should help you answer the first question.
2. (a) Use implicit differentiation to find dy/dx
(b) The line tangent to a function is given by y-y1 = m(x - x1), where m is the slope. They tell you the line is tangent at the point y=1 so plug it in and you will get the x value. Now you have you point (x1,y1). In part (a) u had to find dy/dx which is the derivative, plug x1 into dy/dx which will give you the slope.
3. (a) Just use chain rule.
(b) Find the derivative by using the natural log (ln) properties to get rid of the exponential sin(x).
I see. I posted in your thread. If no one is answering maybe it is because no one on the board. Its Sunday and many people go out etc.
Would you care to post such function that you graphed?
Maybe using calculus you can find the minimum value without having to graph it. Minimum and maximum values occur only at the critical points of the graph where the derivative equals 0 or the derivative doesnt exist or they can also occur at the endpoints of a continuous function in a closed interval.
johnnytheflipper, are you the same as tony1949?
As i said in one of your other posts. Ask for help, dont ask for us to solve your problems. Anyways, im going to solve (a) and tell you how to solve the others like bossk171 did.
Let g(x) = f(x² + 1)
g'(x) = f'(x^2 + 1)(2x)
= 2xf'(x^2 + 1)
You cannot give a final answer without knowing the function f which you didnt specify. g(x) is a composite function of f and x^2 + 1
Now for part b, Just substitute 2 for x.
For part c substitute x^3 for x.
Etc
Is that the graph of g or g' ?
Also, I dont think anyone would like it that you come here and expect us to solve your homework problems. We are here to help you but not do your homework.
Show us that you have tried, what is it exactly that you need help with? Post specific questions.
Could anyone explain the trick of "making the hole larger than the paper"? I do not understand it.
Ok I kind of get it. It is a confusing class, do you agree?
I mean, i remember when I was introduced to Calc and i thought it was hard but now it all seems natural, does it ever get like that with discrete?
Im doing some serious studying today and trying to let everything sink in, you wouldnt mind if I post more questions if i need to?
Thanks a lot for your help!
You just confused me more there... Sorry for bugging you so much but would you care to explain me?
if what the book says is correct then like i said before it leads to conclusions such as this:
p is false, q is true and the conditional statement as a whole is true. If the statement is true it proves that A is a subset of B meaning that (x !e A) and (x e B) and A is a subset of B? How come?
I dont either.
Im going to copy it down just how it is in the book.
"Let p and q be propositions. The conditional statement p -> q is the proposition "if p, then q". The conditional statement p -> q is false when p is true and q is false, and true otherwise. In the conditional statement p -> q, p is called the hypothesis (or antecedent or premise) and q is called the conclusion (or consequence)"
"Note that the statement p -> q is true when both p and q are true and when p is false (no matter what truth value q has)"
What you just said makes PERFECT SENSE.
However, thats not how my book defines "p -> q". In fact, the truth table I gave you is the exact one that goes in my book. In my book it says that given the statement "p -> q", the statement is always true if p is false, and the statement is always true if q is true. It says that literally which didnt make sense to me at all but i just memorized it...
Why is this?
Im SO confused...
Also, thanks for replying.
Please. I would appreciate it if anyone helped me because im kind of confused and I dont want to go to class like this tomorrow.
Theres a definition that says that A is a subset of B iff:
Ax(x e A -> x e B)
is true.
The first A is supposed to be upside down (universal quantifier) and the "e" is the "exists" or "belongs to" symbol.
Now I have a question about this statement. First of all, could anyone please explain me what does the "Ax" mean? I was reading the definition in the book and it says it takes the place of "for every x..." but is that it? Does that symbol just replaces words or does it have a more intricate meaning?
Besides that, that statement could be thought as a conditional statement such as "p -> q" where:
p = x e A
q = x e B
In this sense, if one constructs the truth value of the statement "p -> q" one could see that this statement is true when p is false or when q is true.
Truth table:
p ----- q ----- statement
T ----- F ------ FALSE
T ----- T ----- TRUE
F ----- F ------ TRUE
F ----- T ----- TRUE
Given that (p) -> (q) = (x e A) -> (x e B)
From the truth table one can see that out of the 4 choices, theres 3 choices that yield "true" results. The false results means that in that case A is not a subset of B (we are talking regular subsets, not proper subsets). The other 3 choices yield true which means that given the truth values that p and q takes, A will be a subset of B.
so when p and q are both true, the statement is true which means that for every x in set A theres an x in set B, making A a subset of B.
when p is false and q is false, the statement is true (because p is the hypothesis and when it is false, it makes the statement true) therefore that implies that (x e A) is false and (x e B) is false but since the statement is true, it means that A is a subset of B? How could that be?
This same problem persists with the other 2 truth values....
Would anyone care to explain?
Thanks a lot for your responses guys.
The Discrete Math book we have starts with logic and proofs like you said but the teacher says she likes to start with Set Theory which is the second chapter. I always like to review stuff before and after class with the book but i find that many theorems and definitions of Set Theory are given using stuff from logic and im basically lost...In class she explains everything without using propositions etc but then the book does it differently... Im going crazy.
To be honest, the class looks challenging and it is certainly very different from anything I have done before BUT I think i can do it but certainly the current teaching method of my professor does not help at all...
Would you recommend I read all of Chapter 1 and then get on with Chapter 2 which is what we are doing in class and that way I can understand stuff better?
Im going to try my best on this class.
So this semester im taking Calculus II, Discrete Mathematics and Physics I with Calc...as well as a Enviromental requirement.
First day of class Discrete Math really freak me out. We started with set theory, something that I never seen before. I understood it all but it was certainly something completely different to Calculus.
How hard is this class? Im looking forward to put a lot of effort into but I just want to know what kind of monster im going to dealing with?
Hello guys. Im just back home from taking a Calculus test and although I think I did ok, I missed some questions and wasn't sure about others.
1.) Integrate:
S((x^3+x-2)/(1+x^2))dx
I just didnt know what to do for that so I left it blank.
2.) f(x) = ax^2 +bx + c. Find a,b,c such that f(0)=4 is an absolute minimum of f in [-1,2]
I found that c=4 and b=0 but for the love of god I could not find the value of a....I did however find an inequality that showed that a>0 so I gave my answer as following:
b=0
c=4
a>0.
f(x) = ax^2 + 4, a>0.
That doesnt look like anything we've done in class but I think its better than leaving it blank. I did a couple of tests with different values of a and my answer held true for all values of a. Anyone care to expand on this?
3.) This was a Mean-Value-Theorem with f(x) = sqrt(x) -2x for x in [0,1]. I was able to do the problem completely and get the right answer etc. However, there was a part that asked me to show that f(x) meets all the requirements of the MVT. The requirements of the MVT are that f is continuous on [a,b] and differentiable on (a,b). I just said that f was a polynomial and therefore continuous and differentiable on [a,b] and (a,b) respectively. But I dont think thats right? I dont know, it seems I just pulled an easy answer off my math.
Anyways, I should be fine since I always get the highest grade and it curves up to a 100 so im sure I will still get an A but I just wanted to see the answers to this questions because when i cant do something it bugs me.
Thanks a lot.
Im trying to do this initial-value problem in which I know what to do but I do get a step they do on the Solutions Manual.
Heres the problem
dy/dx=(x^2 - 1)/(x^2 + 1)
y(1) = (pi)/2
Now I know I have to take the integral of dy/dx but in order to take the integral in the Solutions Manual they rewrite dy/dx as:
dy/dx = 1 - (2/(x^2 + 1))
How does one go from the first dy/dx to the rewritten one?
Thanks.
You have some anger control issues. It's supposed to be a game...So one person makes a mistake? No big deal.
Dont jump on me now, im just saying take it easy and lets continue with the game.
Hey Ricky.
I wasn't able to finish it off...Could you help me with that? I never had a formal introduction with proofs so im pretty much clueless...Is there class that teaches you proofs or is that something that you learn by yourself?
I seem to have stumbled upon a problem:
Use the MVT to show that if f is differentiable on an interval I, and
if |f'(x)|<=M for all values of x in I
then
|f(x)-f(y)| <= |x-y|
No idea how to do this...
Yes sorry I forgot f(a)=f(b)
Ahhh I get it. Now if the professor would had said that...
In the context that he explained it, it had no practical use. It just seemed as another formula to memorize. Now I get its purpose.
Thanks guys.