Hey guys,
A few Calculus problems here...
(I've been away from a few classes and am trying to catch up.)
1) Using differentials, approximate sin(46 [sup:1x5tkuw8]o[/sup:1x5tkuw8]) to 3 decimal places.
I used the Newton`s formula x [sub:1x5tkuw8]n+1[/sub:1x5tkuw8] = X [sub:1x5tkuw8]n[/sub:1x5tkuw8] - [f(X[sub:1x5tkuw8]n[/sub:1x5tkuw8]) / f [sup:1x5tkuw8]1[/sup:1x5tkuw8] (X [sub:1x5tkuw8]n[/sub:1x5tkuw8])
setting X [sub:1x5tkuw8]0[/sub:1x5tkuw8] as [ 1/ (square root 2) ] (using sin45[sup:1x5tkuw8]o[/sup:1x5tkuw8] as 'something close to sin 45[sup:1x5tkuw8]o[/sup:1x5tkuw8]' )
I started writing the formula as x[sub:1x5tkuw8]1[/sub:1x5tkuw8] = [ 1 / (square root 2) ] - [ f(X[sub:1x5tkuw8]0[/sub:1x5tkuw8]/ f [sup:1x5tkuw8]1[/sup:1x5tkuw8] (X[sub:1x5tkuw8]0[/sub:1x5tkuw8]) ]
But, what do I put in [ f(X[sub:1x5tkuw8]0[/sub:1x5tkuw8]/ f [sup:1x5tkuw8]1[/sup:1x5tkuw8] (X[sub:1x5tkuw8]0[/sub:1x5tkuw8]) ] ? There's no function for me to put in there, so I am unsure how to proceed from here.
2) For the curve with equation x[sub:1x5tkuw8]3[/sub:1x5tkuw8] + 2y[sub:1x5tkuw8]3[/sub:1x5tkuw8] + 3xy = 0 , find slope of the tangent line at the point (2, - 1).
For this problem, I'm not sure how to even go from here.
3) A water trough with vertical cross-section in the form of an equilateral triangle is being filled at a rate of 4 cubic metres per minute. Given that the trough is 12 metres long, how fast is the level of the water rising when the water reaches a dept of 1.5 metres.
Again, not sure how to start this problem. With no formula to work with, am I supposed to graph it first or something?
Thanks guys!
A few Calculus problems here...
(I've been away from a few classes and am trying to catch up.)
1) Using differentials, approximate sin(46 [sup:1x5tkuw8]o[/sup:1x5tkuw8]) to 3 decimal places.
I used the Newton`s formula x [sub:1x5tkuw8]n+1[/sub:1x5tkuw8] = X [sub:1x5tkuw8]n[/sub:1x5tkuw8] - [f(X[sub:1x5tkuw8]n[/sub:1x5tkuw8]) / f [sup:1x5tkuw8]1[/sup:1x5tkuw8] (X [sub:1x5tkuw8]n[/sub:1x5tkuw8])
setting X [sub:1x5tkuw8]0[/sub:1x5tkuw8] as [ 1/ (square root 2) ] (using sin45[sup:1x5tkuw8]o[/sup:1x5tkuw8] as 'something close to sin 45[sup:1x5tkuw8]o[/sup:1x5tkuw8]' )
I started writing the formula as x[sub:1x5tkuw8]1[/sub:1x5tkuw8] = [ 1 / (square root 2) ] - [ f(X[sub:1x5tkuw8]0[/sub:1x5tkuw8]/ f [sup:1x5tkuw8]1[/sup:1x5tkuw8] (X[sub:1x5tkuw8]0[/sub:1x5tkuw8]) ]
But, what do I put in [ f(X[sub:1x5tkuw8]0[/sub:1x5tkuw8]/ f [sup:1x5tkuw8]1[/sup:1x5tkuw8] (X[sub:1x5tkuw8]0[/sub:1x5tkuw8]) ] ? There's no function for me to put in there, so I am unsure how to proceed from here.
2) For the curve with equation x[sub:1x5tkuw8]3[/sub:1x5tkuw8] + 2y[sub:1x5tkuw8]3[/sub:1x5tkuw8] + 3xy = 0 , find slope of the tangent line at the point (2, - 1).
For this problem, I'm not sure how to even go from here.
3) A water trough with vertical cross-section in the form of an equilateral triangle is being filled at a rate of 4 cubic metres per minute. Given that the trough is 12 metres long, how fast is the level of the water rising when the water reaches a dept of 1.5 metres.
Again, not sure how to start this problem. With no formula to work with, am I supposed to graph it first or something?
Thanks guys!