MHB Solve Force Systems II: Derivatives & Reasoning Explained

  • Thread starter Thread starter Drain Brain
  • Start date Start date
  • Tags Tags
    Force Systems
AI Thread Summary
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.
Drain Brain
Messages
143
Reaction score
0
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!
 

Attachments

  • mechanicsCHII-53.jpg
    mechanicsCHII-53.jpg
    85.2 KB · Views: 115
Last edited:
Mathematics news on Phys.org
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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top