MHB Solve Force Systems II: Derivatives & Reasoning Explained

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The discussion focuses on understanding the reasoning behind taking derivatives in solving a force system problem. The first derivative is set to zero to find the minimum of F_R, and the second derivative test is used to confirm whether this point is a minimum or maximum. Participants clarify that the substitution of F_1=57.8 into the second derivative equation is necessary to evaluate the relationship between F_R and F_1. The calculations lead to a positive second derivative, indicating a minimum point. Overall, the conversation emphasizes the importance of understanding the reasoning behind derivative applications in optimization problems.
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Hello! :)

Here's another problem that I want to fully understand how it was solved.

The part that I'm having a hard time with is the taking-derivatives of some equations. Why did the solver decide to take the derivative of equation 2. And why the second derivative of equation 1 became like that(encircled with red)?
It's the taking-derivatives of things I'm most confused(not the taking derivatives, but the reasoning of the solver why did he take that route.) THANKS!
 

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The problem is looking for the minimum [math]F_R[/math], which requires us to take a derivative. (The derivative of a function is 0 at a relative minimum point.) So we set the 1st derivative to 0. Once that is done we need to see if the value of [math]F_R[/math] given by the 1st derivative is a relative minimum or a relative maximum. The 2nd derivative test does this.

As for the second derivative:
The first derivative equation is:
[math]2F_R ~ \frac{d F_R}{d F_1} = 2F_1 - 115.69[/math]

Taking the derivative with respect to [math]F_1[/math]:
[math]2 \frac{d F_R}{d F_1} \cdot \frac{d F_R}{d F_1} + 2 F_R ~ \frac{d^2 F_R}{d F_1 ^2} = 2[/math]
Now just divide by 2.

(The derivative of the LHS is done by the product rule: [math]\frac{d}{dx} f(x)g(x) = \frac{df}{dx} g(x) + f(x) \frac{dg}{dx}[/math]. Also note that I have taken the derivative on the LHS in a different order than your source so it matches the "usual order" when using the product rule.)

-Dan
 
Hello Everyone! :)

Just want to ask how did the solution arrive at the part where it substitutes $F_{1}=57.8$(which, I suppose the critical point of the first derivative) and $\frac{d F_R}{d F_1}=0$ to the 2nd derivative.

$\displaystyle \frac{d F_R}{d F_1} \cdot \frac{d F_R}{d F_1} + F_R ~ \frac{d^2 F_R}{d F_1 ^2} = 1$ I only see $\frac{d F_R}{d F_1}$ but not $F_{1}$, where I can substitute their values.

which results in

$\frac{d^2 F_R}{d F_1 ^2}=0.00263>0$ --->>> how did it arrive here? I know what this result means, it tells us the point of minimum. But I don't understand how did that happen.

Need an Immediate help here!
 
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Help please! Up! Up! :(
 
Drain Brain said:
Help please! Up! Up! :(
Hi Drain Brain:
In the solution notice that equation (2) gives a relation between FR and F1. Once you have F1, simply find FR using equation (2) and replace into the 2nd derivative relation.
 
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