the tenth week of calculus has been exactly like the first nine. we learned implicit derivatives. and we learnedd the first and second implicit der tests.
First derivative
1) take the derivative
2) everytime you take the derivative of y you note it by saying that it is y prime. You do this by putting one of two things. You can either put dy/dx or y^1.
3) Once you finish the derivative you then solve the equation for dy/dx of y prime. You do this by using simple algebra.
Second Derivative
1) first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.
2) you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.
3) once you have everything plugged in and ready to go you then solve for d^2y/d^2x
This is also done by using simple algebriac methods.
im still not too goood at optimization becuase i always cant find the primary and secondary equations and i never know what to do w/ the missing variables even though i know how to optimize things haha
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im pretty sure all you have to do is solve for a variable and plug back in to the equation but im not sure what after that hope this helps some what
ReplyDeleteokay so the primary equation is what you are maximizing or minimizing and the secondary is the other.
ReplyDeleteA simple problem would be: The product is 192 and the sum is a minimum
Since it tells you the sum is a minimum, you know that is the primary because you are minimizing it and from there you follow the steps you said you know how to do
A harder example would be: A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with MAXIMUM VOLUME.
You know you are maximizing the volume therefore v=x^2h is the primary formula and Surface area would be the secondary formula.