Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Defined integral, but one limit needs to be found 2

Status
Not open for further replies.

AnaAnt

Civil/Environmental
Mar 24, 2012
41
Hi guys!
Need a little help couse Mathcad doesnt allow me to deine the integral limit as the unknown.
Simple egsample:

integral of "x dx" with limits "a=0 and b=5".
The solution is 12.5
How to reverse the situation so that a=0, b=unknown and provide the previous result of 12.5?
I want Mathcad to find the b value.
How to do this?
Thanx
 
Replies continue below

Recommended for you

That should be doable with a solve block

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
"solve block"
Is that a command inside Mathcad or something else?
 
...and you like to drink wich beer and how much? :)
 
One more question, and thenx for the previous answer.
How to use the result you get?
Im asking this becasue if I have even the simpliest equation like 2x+3=0, in Mathcad I would get -3/2 using the "solve for x".
But how to use the x value (-3/2) for your further calculations?
 
Root functions seems to work for this couse I can get numeric value.
Anyother ideas?
 
"But how to use the x value (-3/2) for your further calculations?"

That would depend on what you're doing, wouldn't it? Presumably, x is something you needed. If you didn't need the solution, then there wouldn't be any point in solving for it, would there?

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
Sorry I dont understand you :-(
Something on "root" function?
 
I don't understand why you say that getting the solution like -3/2 is useless because it's somehow not "numeric?"

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
No, I think he means -->solve,x and it doesn't work.
AnaAnt: What you have to do is --- x:=1 (or something) then Given [CR] 2*x+3:=0 [CR] x:=Find(x). After Given you have to use [CRTL]= just like before.
 
Hi guys.
One more question on root function in MATHCAD.
Plase look at the file attached.

When I try to calculate t(x)=0 and find the the x value "by hande" I get an quadriatic equation with x1 and x2.
Using the root function in Mathcad I always get warning message saying "divide by zero" :-(
What am I doing wrong?
Thank you.
 
 http://files.engineering.com/getfile.aspx?folder=6c6a6a22-4c80-41e9-a3ac-5d4efd5f8453&file=rootfunction.xmcd
mar2805 said:
I always get warning message saying "divide by zero" sad
What am I doing wrong?
You tell Mathcad to search for a solution in the range from 0 to h.
Look at your function - what happens for x=0? (Tip: look inside the square brackets ;-)).
So replace the 0 for something like 10^-2 mm (you need to add the unit) and you get 135.017mm.
As an alternative you can use root another way:
x:=1m // guess value needed
z:=root(t(x),x)
 
PNachtwey said:
I typed integrate(x^2,0,b)=12.5,solve,b and it returned 5 and -5.
Thats one of the possible ways to answer the original question which I posted here 12 days ago. See above.
Was this a question of yours?
 
Quote (PNachtwey)
I typed integrate(x^2,0,b)=12.5,solve,b and it returned 5 and -5.
Just saw it - I hope you didn't really type x^2, otherwise this would mean you have a corrupt copy of mathcad ;-)
 
I am having a similar issue in that I use a solve block to solve for two unknowns say C1 & C2. I use the Find(C1,C2) and it gives me the values in a 1x2 Matrix form. After the solve/find block I would like to use C1 and C2 in further equations but they are not defined. I have to manually set C1= to the value I see in the find block. How do I define C1 & C2 as the values in the Find block?

Occupant above gave an example that works when there is one undefined variable ie. C1 := Find(C1) but I can't get it to work with 2 unknowns.
Tried something like this C1,C2 := Find (C1,C2) ..... or defining them separately C1 := Find(C1) C2 := Find(C2). It wont solve when I break it up.
 
Shinerttu…

You can extract values from vectors and matrices by using array subscripts. For example (and asusming ORIGIN=0), if you have Res:=Find(C1,C2)=[C1 C2], you can get C1 out of the result vector by using Res[0=[/b] and you can get C2 by using Res[1=[/b]. The "[" keystroke gets you the array subscript. If ORIGIN=1, then use 1 and 2 as your array subscripts.

==========
"Is it the only lesson of history that mankind is unteachable?"
--Winston S. Churchill
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor