simplifying a series of equations

Questions
1

Hi
I have a series of interconnected equations that I know will simplify to a polynomial but I have been unable to do so. Is this possible?

y=t-DT

where

DT = 300a/9.81

where

a=[(B/2)^2/3]/[(Tu/2)^2/3 * (h-hv+tu)]

where

hv = (Vi * d)/(4B * (Tu/2)^2)^1/3

Where

B = Vi * g * d

where

g = (9.81 (t-x))/300

Vi = Q/l

I want to express this in the terms of

y = ax^2 + bx + c

where a, b and c will be some combination of the variables detailed above

2

Hi Louis,

Making your substitutions,

y==t-DT /.DT->300a/9.81
y==-30.58103975535168` a+t/.a->((B/2)^2/3)/((Tu/2)^2/3*(h-hv+tu))
y==t-(30.58103975535168` B^2)/((h-hv+tu) Tu^2)/.hv->(Vi*d)/(4B*(Tu/2)^2)^1/3
y==t-(30.58103975535168` B^2)/(Tu^2 (h+tu-(d Vi)/(3 B Tu^2)))/.B->Vi*g*d
y==t-(30.58103975535168` d^2 g^2 Vi^2)/((h+tu-1/(3 g Tu^2)) Tu^2)/.g->(9.81 (t-x))/300
y==t-(0.032700000000000014` d^2 Vi^2 (t-x)^2)/(Tu^2 (h+tu-10.193679918450558`/(Tu^2 (t-x))))/.Vi->Q/l
y==t-(0.032700000000000014` d^2 Q^2 (t-x)^2)/(l^2 Tu^2 (h+tu-10.193679918450558`/(Tu^2 (t-x))))

So, the final form of y is:

y==t-(A (t-x)^2)/(B-F/(t-x))

This cannot be put into a quadratic form. If you solve for x, there would be two solutions if it were quadratic:

In[21]:= Solve[y==t-(A (t-x)^2)/(B-F/(t-x)),x]

Out[21]= {

{x->t+(2^(1/3) (-3 A B t+3 A B y))/(3 A (-27 A^2 F t+27 A^2 F y+\[Sqrt](4 (-3 A B t+3 A B y)^3+(-27 A^2 F t+27 A^2 F y)^2))^(1/3))-(1/(3 2^(1/3) A))((-27 A^2 F t+27 A^2 F y+\[Sqrt](4 (-3 A B t+3 A B y)^3+(-27 A^2 F t+27 A^2 F y)^2))^(1/3))},

{x->t-((1+I Sqrt[3]) (-3 A B t+3 A B y))/(3 2^(2/3) A (-27 A^2 F t+27 A^2 F y+\[Sqrt](4 (-3 A B t+3 A B y)^3+(-27 A^2 F t+27 A^2 F y)^2))^(1/3))+(1/(6 2^(1/3) A))(1-I Sqrt[3]) (-27 A^2 F t+27 A^2 F y+\[Sqrt](4 (-3 A B t+3 A B y)^3+(-27 A^2 F t+27 A^2 F y)^2))^(1/3)},

{x->t-((1-I Sqrt[3]) (-3 A B t+3 A B y))/(3 2^(2/3) A (-27 A^2 F t+27 A^2 F y+\[Sqrt](4 (-3 A B t+3 A B y)^3+(-27 A^2 F t+27 A^2 F y)^2))^(1/3))+(1/(6 2^(1/3) A))(1+I Sqrt[3]) (-27 A^2 F t+27 A^2 F y+\[Sqrt](4 (-3 A B t+3 A B y)^3+(-27 A^2 F t+27 A^2 F y)^2))^(1/3)}
}

But there are three solutions.

Eric

3

Try this

In[14]:= y = t - DT; DT = 
 300 a/9.81; hv = (Vi*d)/(4 B*(Tu/2)^2)^(1/3); B = 
 Vi*g*d; g = (9.81 (t - x))/300; Vi = Q/l;

In[15]:= y

Out[15]= -30.581 a + t
4

thanks for your reply Eric it is much appreciated

I am new to this website, so am just trying to learn how it all works. Did you calculate the substitutions yourself or do you use wolfram?

5

If you include the equation for a:

y=t-DT;
DT=300 a/9.81;
a=((B/2)^2/3)/((Tu/2)^2/3*(h-hv+tu));
hv=(Vi*d)/(4 B*(Tu/2)^2)^(1/3);
B=Vi*g*d;
g=(9.81 (t-x))/300;
Vi=Q/l;

In[34]:= y
Out[34]= t-(0.0327 d^2 Q^2 (t-x)^2)/(l^2 Tu^2 (h+tu-(3.12716 d Q)/(l ((d Q Tu^2 (t-x))/l)^(1/3))))

This doesn’t agree with my substitutions, though. ?

Louis, Mathematica did all the work.

Eric

Edit:

The error is mine; I didn’t put a pair of parentheses around the 1/3 power in hv. So are the other powers 2/3?

Louis, could you more completely parenthesize the equations? The result could be very different.