Thanks excognito for all of your help & I will look at your notes when I get home. You have been a great help.
jghrist that is an interesting use of the first derivative root for me.
Thank you, I'm looking into the points you raised.
I've attached a graph of the function V(x). The maximum value on it shows 14.993. Can a 2nd derivative test prove this do you think?http://files.engineering.com/getfile.aspx?folder=73dc5224-aa4f-48db-9d11-91124ef669b6&file=Engtips_1.mcd
Right I was okay up until the last bit & these are probably daft questions but you have helped me immensely so far:
What is the answer to the max volume as there are 3 values?
-10
10.657 (zero y value)(root)
-0.657 (zero y value)(root)
When you write "V(roots[0)=" what is the [ for?
Also...
So the total volume was V(x)=(10+x)/6.sqrt(32-(5-x)^2) with x between 0 & 10.657, how do you get the maximum volume?
Do I plug in values for x within the range given?
Wow again! I'm learning so much just in 1 question!
Why isn't a solve block suitable here?
So do I understand you that roots returns a value of x when the f(x)=0
Im is also a new operator to me.
Okay so if I wanted to find a value of x where V'(x)=0 I presume I use a solve block.
If so would I now write:
dVs2(x):=0
Given
then put the dVs2(x) equal to something?
Hi,
Can anyone take a look at my file to see if I have understood the what is being asked of me properly as I have only just started Mathcad & need all the help I can get!
Thank you
