Okayy so this week we basically did what we always do..
took AP's and corrected them.
i'm just going to go over some stuff we see on every test.
A particle's position is given by s=t^3-6t^2+9t what is the acceleration at time t=4
t^3-6t^2+9t
take deriv: 3t^2-12t+9
take second deriv: 6t-12
plug in 4: 6(4)-12= 12
Original - Position
1st Derivative - Velocity
2nd Derivative - Acceleration
VOLUME:
y=the sqr. root of (-2(x)^2-10(x)+48) inbetween 1 and -2
graph, then put the equation they give you into the integral and square it
PI times the integral of (-2(x)^2-10(x)+48) dx
PI [(-2/3)(x)^3-5(x)2+48(x)] of 1 and -2
f(1)=(127/3) f(-2)=(-332/3)
ANSWER: 153PI
The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
questions: the first 3 problems on the calculator portion=helppp.
umm a review on substitution would also be nicee.
substitution:
ReplyDeletefind your u and then take the derivative of that for du
then figure out what's missing and deal with that first...
next do like usual just with u and du in the problem instead of x and dx and other numbers
then simplify and plug back in the u [and du if needed]
simplify and that's your answer!