Quotient rule with additional condition

In summary, Homework Statement states that the student is working on a chemical reaction engineering problem and it involves some math. They have to find the residence time for maximum production and they have to find the τ (residence time). The Attempt at a Solution states that the student has to use the quotient rule and they figure out that d[f/g]=(f'g-fg')/g^2. When they use the rule to solve for τ, they come up with an incorrect answer. However, if they substitute g^2=(1+0.4τ+0.04τ^2)^2 into the equation, they get the correct answer. The next step would be to try to find a
  • #1
Fluidman117
34
0

Homework Statement



I am working on chemical reaction engineering problem and it involves some math, which I am not able to figure out...

I have to find the residence time for maximum production, which is in the case when : (dη_p)/dτ=0

I have to find the τ (residence time).

Homework Equations



η_p=0.1τ/(1+0.4τ+0.04τ^2)

The Attempt at a Solution



I figure I have to use the quotient rule:

d[f/g]=(f'g-fg')/g^2

f=0.1τ
g=1+0.4τ+0.04τ^2

When I use the rule, I come up with an number which is not the same as in the answers. The reason is I think that I don't know how to take into account the condition that (dη_p)/dτ=0
 
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  • #2
The derivative has to be zero, that means an equation for τ, easy to solve. Show your further work please.

ehild
 
  • #3
When you say the derivative equals zero, do you mean this:

0=(f'g-fg')/g^2

0=[0.1(1+0.4τ+0.04τ^2)-(0.1τ)(0.4+0.08τ)]/g^2
0=0.1-0.004τ^2/g^2

0.1-0.004τ^2=0

and solving for τ=5 , which is the correct answer.

However, I don't understand why g^2=0 ?
 
  • #4
g^2 is NOT 0. What makes you think it is?
 
  • #5
I don't have any scientific reasoning behind my logic, but basically that is the only way I could see that I come up with the answer given in the book.If I subsitute g^2=(1+0.4τ+0.04τ^2)^2 and using the formula
(a+b+c)^2= a^2+b^2+c^2+ 2(ab+bc+ca)

I end up with

0=(f'g-fg')/g^2=(0.1-0.004τ^2)/(1+0.8τ+0.24τ^2+0.032τ^3+0.0016τ^4)

What would be the next step from here?
 
  • #6
Fluidman117 said:
When you say the derivative equals zero, do you mean this:

0=(f'g-fg')/g^2

0=[0.1(1+0.4τ+0.04τ^2)-(0.1τ)(0.4+0.08τ)]/g^2
0=0.1-0.004τ^2/g^2

0.1-0.004τ^2=0

and solving for τ=5 , which is the correct answer.

However, I don't understand why g^2=0 ?

If g^2 = 0, the derivative is undefined. The quantity (f'g - fg') must be 0 to satisfy the equation.
 
  • #7
SteamKing said:
If g^2 = 0, the derivative is undefined. The quantity (f'g - fg') must be 0 to satisfy the equation.
To elaborate on what SteamKing said, the basic idea is that if a/b = 0, then a = 0. Of course, b cannot be 0.
 
  • #8
Thanks everyone!
 

Related to Quotient rule with additional condition

1. What is the Quotient Rule with an additional condition?

The Quotient Rule with an additional condition is a mathematical formula used to find the derivative of a quotient function when the denominator is not equal to zero. The additional condition refers to the fact that the derivative can only be found if the denominator is not equal to zero.

2. Why is the additional condition necessary?

The additional condition is necessary because the derivative of a quotient function is undefined when the denominator is equal to zero. This is because division by zero is mathematically undefined. Therefore, the additional condition ensures that the derivative is only computed when it is defined.

3. How is the Quotient Rule with additional condition different from the regular Quotient Rule?

The regular Quotient Rule can be used to find the derivative of a quotient function in general, without any restrictions. However, the Quotient Rule with additional condition takes into account the restriction that the denominator must not be equal to zero. This means that the derivative can only be found under this additional condition.

4. What is the benefit of using the Quotient Rule with additional condition?

The benefit of using the Quotient Rule with additional condition is that it allows us to find the derivative of a quotient function even when the denominator is not a constant value. This makes it a more versatile tool in solving mathematical problems involving quotient functions.

5. Can the Quotient Rule with additional condition be applied to all quotient functions?

No, the Quotient Rule with additional condition can only be applied to quotient functions where the denominator is not equal to zero. If the denominator is equal to zero, the derivative cannot be found using this rule. In such cases, other rules or methods must be used to find the derivative.

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