|
not asking for answers!
1.
answer: sqrt(x^2+5)+C
2.
answer: (6-x^2)/ 2sqrt(x)(x^2+6)^(3/2)
3.
answer: 9xsec sqrt(9x^2+7)^2 / sqrt(9x^2+7)
4.
answer: okay this im unsure but ill show you the steps.
y = sin2x x = pi/4
f(x) = sin2(pi/4) = 1 this is y.
f'(x) = 2cos(2x) = 0 this is m
y = mx + b
1 = 0x + b
1 = b
y = mx + 1 <-- is this right?
5.
answer: 5cos(8x^5) - 200x^5sin(8x^5)
the numbers im very unsure about are 3 & 5, i mightve screwed up a step(s). im pretty sure about #4 but is that a valid answer? just asking to check and not begging for answers.
|
United States24698 Posts
Can't you verify stuff like this easily with math software without bugging anyone? They are all just derivatives and crap...
|
i cant do that for tests though. 
|
We can't verify your test answers either.
|
1 is right, except that you found them all and not just one 
Essentially, you can just make c whatever constant you want, 0 is easy.
2. I'm not so sure about that one, I got a different answer, which agreed using two methods.
3. Is close, what is the derivated of tanx?
4. y=mx+1 is right if m=0, in which case it is just the horizonal line y=1. But you should be able to check that graphically. Sketch sin2x and look at what the tangent will look like at pi/4!
5. Looks good.
|
On October 24 2008 12:09 conCentrate9 wrote:
We can't verify your test answers either.
no shit einstein, but this gives me better chances to eliminate mistakes.
|
FuDDx
United States5008 Posts
Let us know how WE scored on YOUR test.^_^
|
f u guys! close thread. f u guys!!!!!1
|
|
|
United States4991 Posts
http://integrals.wolfram.com/index.jsp
Just use that if you need to check your integration answers.
ed: I realize you're finding derivatives btw, but you can still use it to check your answers.
ed2: I realize also that micronesia suggested something like that before, but I provided an actual link 
|
On October 24 2008 12:10 Zortch wrote:1 is right, except that you found them all and not just one Essentially, you can just make c whatever constant you want, 0 is easy. 2. I'm not so sure about that one, I got a different answer, which agreed using two methods. 3. Is close, what is the derivated of tanx? 4. y=mx+1 is right if m=0, in which case it is just the horizonal line y=1. But you should be able to check that graphically. Sketch sin2x and look at what the tangent will look like at pi/4! 5. Looks good.
i dont get what you meant by 1 though, i mean. if it doesnt have "0 to 4" in integration, wouldnt there always be a +constant?
2. yeah this one is tricky, i dont know. what did you get for this? you might be right and i could be wrong.
3. tanx is 1/cosx^2
thank you!! :D
|
FuDDx
United States5008 Posts
Im joking sir calm yourself ^_^ honestly have no clue about any of this ill see myself out thanks.
|
Yea so for 1, the question is, find a function g s.t. g' =whatever.
So there are infinetly many of these since any constant will do.
g(x)=sqrt(x^2+5)+C for any c - a real number (or whatever field you're in but don't worry about that)
For example, c could be 1 then, g(x)=sqrt(x^2+5)+1 works, or as i suggested,
g(x)=sqrt(x^2+5)+0=sqrt(x^2+5) works too.
If you're calculating an indefinite integral as you suggested then yes, the +c is required, however in this case you were only asked for one function and there are many choices, thats all  .
For 2, the quotient rule is a pain in the ass imho, so I'd reccomend using the product rule.
You've got sqrt(x/(x^2+6)), but that is equal to
sqrt(x)/sqrt(x^2+6)
=(x^1/2)/((x^2+6)^1/2)
=(x^1/2)((x^2+6)^-1/2)
then use the product rule and see what you get.
My answer:
+ Show Spoiler +(1/2)x^(-1/2)((x^2+6)^(-1/2))-x^(3/2)((x^2+6)^(-3/2)) it could be simplified a bit, but w/e
For 3, that is right, so I think in your answer you just took the derivative of tan to be sec instead of sec^2, maybe just a typo?
Edit: Oh I seee for 3 now. You have:
9xsec sqrt(9x^2+7)^2 / sqrt(9x^2+7)
Should be:
9x (sec sqrt(9x^2+7))^2 / sqrt(9x^2+7) just for clarity, i assume you have it right.
|
I think all of the answers are correct. The answer for four is y=1, as m=0. The answer to 3 is right, but the syntax is confusing.
|
1. correct if you're finding the general integral for any constant. otherwise, i'm too lazy to read through Zortch's... what Zortch wrote seems alright.
2. i got something different, do it once more over and see if you get something different too.
3. as Zortch said, it's the derivative of tanx that is the root of the problem.
4&5 I'm too lazy to go through in detail they look alright.
|
so zortch for 1, my answer will do or?
for number 2 yeah, quotient rule was what i used, its pretty sloppy too. i did chain rule then quotient.
for 3,
9xsec sqrt(9x^2+7)^2 / sqrt(9x^2+7)
9x (sec sqrt(9x^2+7))^2 / sqrt(9x^2+7) yeah its more clear your way, its hard to type the answer you written so it was my fault sorry.
question for 4, can it just be y=1 since mx is 0?
|
For 1, your answer is probably fine, unless they really throw the book at you.
Do you understand what I was saying though?
Yea for 2, stuff like that were you have a quotient to an exponent, its often a good idea to take it at the numerator^exp/denomonator^exp and then use quotient or product rule.
For 4, yea whats wrong with y=1? Thats a line, y=mx+b is a standard form of a line, but if m=0 then m*x=0 so just put 0, same goes for b. If b=0 just put in zero heh.
Sketch sin2x!!
Look at what the tangent line at pi/2 looks like, this is super important and a very useful thing to be able to do.
|
Number 2:
Product Rule -
d/dx [X^0.5 * (x^2+6)^(-0.5)] = (1/2)(x^(-0.5)(x^2+6)^(-0.5) + (-1/2)(x^0.5)(x^2+6)^(-3/2)(2x)
=(1/2)(6-x^2) / [(x^0.5)(x^2+6)^(3/2)] = (6-x^2)/ 2sqrt(x)(x^2+6)^(3/2)
Quotient Rule -
d/dx (x/(x^2+6))^0.5 = (1/2)(x/(x^2+6))^(-1/2)(x^2+6-2x^2)/(x^2+6)^2 = (6-x^2)/ 2sqrt(x)(x^2+6)^(3/2)
I think the answer is correct.
|
On October 24 2008 13:20 Zortch wrote:
For 1, your answer is probably fine, unless they really throw the book at you.
Do you understand what I was saying though?
Yea for 2, stuff like that were you have a quotient to an exponent, its often a good idea to take it at the numerator^exp/denomonator^exp and then use quotient or product rule.
For 4, yea whats wrong with y=1? Thats a line, y=mx+b is a standard form of a line, but if m=0 then m*x=0 so just put 0, same goes for b. If b=0 just put in zero heh.
Sketch sin2x!!
Look at what the tangent line at pi/2 looks like, this is super important and a very useful thing to be able to do.
yeah i do, but im afraid it would be wrong because someone told me that its asking "a" so the +c isnt needed and its just the answer given.
On October 24 2008 13:21 Descent wrote:
Number 2:
Product Rule -
d/dx [X^0.5 * (x^2+6)^(-0.5)] = (1/2)(x^(-0.5)(x^2+6)^(-0.5) + (-1/2)(x^0.5)(x^2+6)^(-3/2)(2x)
=(1/2)(6-x^2) / [(x^0.5)(x^2+6)^(3/2)] = (6-x^2)/ 2sqrt(x)(x^2+6)^(3/2)
Quotient Rule -
d/dx (x/(x^2+6))^0.5 = (1/2)(x/(x^2+6))^(-1/2)(x^2+6-2x^2)/(x^2+6)^2 = (6-x^2)/ 2sqrt(x)(x^2+6)^(3/2)
I think the answer is correct.
haha wow, our maths are different that mine is longer and messier. yeah i cant skip steps, on the net i do bc im lazy but when im writing i need every step imaginable. thanks so much.
|
oh for number 5, lets say i would just write it like this
5cos8x^5 - 200x^5sin8x^5
would it be correct or it wouldnt and the parenthesis is really needed?
|
|
|
|
|
|
EPT![[image loading]](http://i34.tinypic.com/15gsdgz.jpg)
![[image loading]](http://i34.tinypic.com/ao8k6r.jpg)
![[image loading]](http://i33.tinypic.com/11j5bf5.jpg)
![[image loading]](http://i34.tinypic.com/x0r1b5.jpg)
![[image loading]](http://i36.tinypic.com/al1yra.jpg)
