yeah, so this week we took tests every single day, and had the career day thing friday, and the tests were super hard, but hopefully i learn from my mistakes and do better etc... so anyways, here is my blog
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1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
Example: the variables x and y are functions of t and related by the equation y=2x^3-x+4 when x=2, dy/dt=-1. Find dy/dt when x=2
alright, so you put down the equation, y=2x^3-x+4.
Then you take the derivative of that, so you get dy/dt=6x^2(dx/dt)-(dx/dt)
then you plug in to find that dy/dt=6(2)^2(-1)-(-1)
and that is further simplified to, dy/dt=-23.
Linearization:
f(x)=f(c)+f'(c)(x-c)
example: Approximate the tangent line to y=x^2 at x=1
you find all the different values: dy/dx=2x dy/dx=2 y=(1)^2=1
then you plug into the formula to get: f(x)=1+2(x-1)
example 2: use differentials to approximate: sq root(16.5)
steps:
1: identify an equation--- f(x)=sq root(x)
2:f(x)+f;(x)dx--- sqrt(x)+ (1/(2sqrt(x)))(dx)
3:determine dx-- .5
4:determine x--- 16
5:plug in--- sqrt(16)+(1/2sqrt(16))(.5)= 4.0625
error= .0005
yeah, so um, i guess i could use some help on taking the second implicit derivative of something, i always manage to screw it up somehow..
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I usually mess up on that too haha. All I can say is that make sure to put d^2y/dx^2 when taking the derivative of dy/dx and still follow all the same steps as usual.
ReplyDeleteyeah second implicit derivative...the most important thing to notice here is that you will almost always use your first implicit derivative. So, solve for the implicit derivative...then proceed to take the derivative again. This time, whenever you get dy/dx, however, you will plug in whatever you found your first implicit derivative to be.
ReplyDeletewhen using the second implicit you have to remember the first implicit and just take the derivative again.
ReplyDeleteTake the first derivative, solve for dy/dx, then plug in your xs and ys to get a number. Take the derivative of dy/dx again and wherever there is a dy/dx in your second derivative, plug in your number you got when you plugedin your xs and your ys.
ReplyDeleteTake your first implicit derivative regularly. Then when taking the second derivative, remember to plug in what you got for the first implicit derivative where dy/dx is. Don't forget to keep it solved for d2y/d2x.
ReplyDeleteTake the derivative of the function like you would a normal dy/dx problem. Once you have solved for dy/dx, plug in the x and y values in to get a value for dy/dx. Then take the derivative of the dy/dx function again, and this time plug in the x, y, and dy/dx values to get the final answer.
ReplyDelete