So, the twelfth week of school and calculus is over, but we didn't really go over any new topics this week. We reviewed and studied for the tests on wednesday and thursday, and i understand related rates and implicit derivatives and second implicit derivatives, but some of the stuff on those tests were super hard! but a lot of it was my fault because i did not study good, next time, i'll remember to study rolles theorem, mean value theorem, and all those geometric formulas that come in handy for related rates. anyway, i supposed i should do an example problem or two, to show what i do understand. here we go...
first, i'll say some stuff about limits that i failed to know before we took the tests.
1. if the top and bottom exponents are the same, the answer is the top coefficient over the bottom coefficient.
2. if the top exponent is bigger than the bottom exponent, the answer is infinity.
3. if the top is less than the bottom, it goes to 0.
Next, i just found a problem that i knew how to do in some old packet that we had, so i will post this as an example.
EXAMPLE: Assume that x and y are both differentiable functions of t. Find dy/dt when x=64 and dx/dt=5 for the equation y=sqrt(x)
first, you take the derivative of the equation, so you get dy/dt=1/2(x)^-1/2(dx/dt)
then, you just plug in the given values, so you get dy/dt=1/2(64)^-1/2(5)
and when you solve, you get dy/dt=5/16.
now, for what i don't understand, it is linearization, big surprize... i know you have to draw the little picture, and take the derivative of the formula, and plug in and everything, but i just don't know how to do the stuff...
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When a problem asks for linearization, they will always have the word approximate in the problem.
ReplyDeleteSo let's say they ask you to approximate the tangent line to y=x^4 at x=2
your dy/dx would be 4x. Then you plug in the x value 4(2) and come out with dy/dx=8.
Then plug in you the x value into the function
y=(2)^4
y=16
so the tangent line would be:
f(x)= 16+8(x-16)
hope this helps!
In linearization the key word is approximate.
ReplyDeleteThe type we saw the most so far is use differentials to approximate something.
For these:
1. identify an equation
2. f(x)+f'(x)dx is the equation you use
3. determine dx
4. determine x
5. plug dx and x into the equation
Example:
use differentials to approximate the square root of 36.8
1. the equation: the square root of x
2. to find f'(x), you have to take the derivative of the square root of x, which is 1/2 square root of x
so your equation is now the square root of x + 1/2 square root of x dx
3. dx= .8
4. x= 36
5. plug in: the square root of 36 + 1/ 2 the square root of 36 (.8)
when figured out the answer is: 6.0667
then find the error which is .0003