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1. 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
y=2x^3-x+4
take deriv: dy/dt=6x^2(dx/dt)-(dx/dt)
plut in: dy/dt=6(2)^2(-1)-(-1)
dy/dt=-23
2. given f(x)=2x^2-7x-10, find the absolute maximum of f(x) on [-1,3]
f(x)=2x^2-7x-10
4x-7=0
=7/4
2(7/4)^2-7(7/4)-10= -129/8
2(-1)^2-7(-1)-10= -1
2(3)^2-7(3)-10= -13
answer is -1 for absolute max
3. chain rule-work from outside in
sin^2(x^2)
2(sin (x^2))(cos(x^2)(2x)
4xsin(x^2)cos(x^2)
4. average value of f(x)=1/x from x=1 to x=e is
1/e-1[ln (absoulte value of e)- ln (absoulte value of 1)]
1/e-1[1-0]
1/e-1
5. 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
i need help on something like
d/dx (integral from 2x to 5x)cost dt
and
like number 44 on the last calculator portion about a radioactive isotpe, y, decays..
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okay...so for the d/dx (integral from 2x to 5x)cos(t)dt
ReplyDeletethe d/dx and the integral cancel each other out
but since is the integral from 2x to 5x they have to be plugged in...usually however, you just plub in the one that's not 0 or the one with an x in it...
the answer should be:
cos(5x)-cos(2x)
For the problem d/dx integral of cos(t) dt from 2x to 5x, you just plug in the bouds. Since they're asking you to take the derivative of an integral,these two cancle out. After that, you just plug in your bounds as you usually would. Of course you won't get a number, you'll get an equation because there are xs in the bounds. So you'll get something like
ReplyDeletecos(5x) - cos(2x)