Solve Force Systems II: Derivatives & Reasoning Explained

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SUMMARY

The discussion focuses on the process of taking derivatives in the context of solving force systems, specifically for finding the minimum of the function F_R. The first derivative of F_R is set to zero to identify critical points, while the second derivative test is employed to determine whether these points are minima or maxima. The participants clarify the reasoning behind substituting specific values, such as F_1 = 57.8, into the second derivative equation to arrive at the conclusion that the second derivative is positive, indicating a minimum point.

PREREQUISITES
  • Understanding of calculus, specifically differentiation techniques.
  • Familiarity with the product rule for derivatives.
  • Knowledge of critical points and their significance in optimization problems.
  • Basic grasp of force systems and their equations.
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  • Study the application of the product rule in calculus.
  • Learn about optimization techniques in calculus, focusing on first and second derivative tests.
  • Explore the relationship between force systems and their equations in physics.
  • Investigate how to derive and interpret critical points in multivariable functions.
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Students and professionals in physics and engineering, particularly those involved in mechanics and optimization problems, will benefit from this discussion.

Drain Brain
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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!
 
Last edited:
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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