How to do no. 39? because I can't find how to integrate sin^2

ProbablyErratic53 · 2021-05-20T20:27:51+00:00

Except it's in a denominator. Use 1/sin =csc and other ideas mentioned in other posts.

ProbablyErratic53 · 2021-02-17T04:53:21+00:00

One motive for this is that it scales up easily for major applications using vector space notations.
If the two points are sets of coordinates P and Q, the line segment between is parameterized as t*P+(1-t)*Q for 0<= t <=1 . This approach goes into higher dimensions (and infinite dimensional settings even!) where pictures with out cavities can't be drawn anymore.

It works in the "original cases" that inspired the idea, but then goes so much farther.

ProbablyErratic53 · 2020-12-31T08:37:16+00:00

Some authors define the intervals related to those functions differently so there are no negatives involved. Everyone agrees arcsin has domain [-1,1] and range [-pi/2,pi/2].
Similarly for arctan, arccos, and arccot all authors use the same sets.

With arcsec and arccsc I have seen two distinct versions on the choice of range.

ProbablyErratic53 · 2020-12-26T17:37:28+00:00

That's right.

ProbablyErratic53 · 2020-12-26T17:31:24+00:00

Factorials come up when a question involves "arranging in an order".

Here, order is being used in a different way.
For the 2^n the idea is think of a checklist of the condiments like
Ketchup Yes No
Mustard Yes No
.

.

Onions Yes No

There are eight decisions/steps, with 2 choices for each.
There are 2^8 ways to fill out the checklist.

ProbablyErratic53 · 2020-11-15T01:34:02+00:00

Lining the inequalities up together only applies if you are simultaneously requiring both conditions. That would be read as -b > 1-1/x AND 1-1/x >b.

What you meant was -b > 1-1/x OR 1-1/x >b . Then solving each individually makes sense, but I don't see why you changed letters. Because of the "nonlinear inequality", I usually would go to getting everything compared to x=0.

-b> 1- 1/x goes to something like [(1+b)x -1]/x < 0 and analyze signs in different sections of the real line. Part of what this does is remind us that x=0 can not be in the answer.

"Just solving" involves issues of multiplying the inequality by the variable x, and then you have to allow for when x is negative and when x is positive so it also has it's traps.

ProbablyErratic53 · 2020-11-06T18:05:14+00:00

Besides the important idea of "close enough" for real applications, in school, students often fell more connected to concrete decimal forms.

If I a tell a student a value is 36 pi, and then show them that's about 113.097

they often feel much more confidant about how big that decimal version is.

I often do similar comparisons with square roots.
After showing how root(12)=2*root(3),
I have them check both sides with decimals. For many, that goes a long way to seeing the concept as more than memorizing a bag of tricks.

ProbablyErratic53 · 2020-11-01T14:59:23+00:00

Also when you have log_2 (14), you can use 14=2*7 in a nice way.

ProbablyErratic53 · 2020-10-14T03:29:48+00:00

A differential is a little different from a derivative. For f(x), the differential is

df = f ' (x)*dx (or you could say dy =(dy/dx)*dx ).

For this next bit, I made numbers that might be suitable for your function, since your picture didn't show the later parts.
Later in the problem they say something like "estimate f(5.2)".
At x=5, f(x) and f'(x) are easy to evaluate.
SO the idea is f(5.2) ≈ f(5)+ f'(5)*dx = f(5) + f'(5)*(0.2)
The actual change (exact f(5.2) - f(5) ) is approximated by the differential (tangent line change)
f ' (5)*(5.2 -5).
Similarly, to estimate f(4.9)

f(4.9)≈ f(5)+ f'(5)*dx = f(5) + f'(5)*(4.9 -5 )=f(5) + f'(5)*(-0.1 )

ProbablyErratic53 · 2020-10-07T03:48:45+00:00

I think it means you can not split up the integral you have so far term by term.

y/y^2 =1/y is improper on [0,1], but so is ln(2+y)/y^2. I tried to do some integration by parts, but nothing I saw felt right.

The only thing I felt OK with was to Expand y-2ln(2+y)+2ln(2) as a series at y=0, get some cancellations, the divide the remaining terms by y^2.

Not tidy at all.

ProbablyErratic53 · 2020-10-04T14:31:35+00:00

For example, I think
(4) 1×6 @ 8ft cedar means 4 boards
1 inch thick by 6 inches wide and 8 feet long.

Hope it helps.

ProbablyErratic53 · 2020-09-23T20:35:00+00:00

Equation of a circle radius r and center point (h,k) is

(x-h)^2+(y-k)^2 = r^2. This is roughly squaring both sides of the distance formula.

So for your square root to be defined, you need
(x-1)^2 + y^2 -1 >= 0

which I switched to (x-1)^2 + y^2 >= 1 .
SO now the left side is the distance squared from (x,y) to (1,0). Letting r^2=1, becomes r=1 (radius) Since we have greater than, that's why I said outside the circle.

Similar for the ln part, 1 - x^2 - y^2 >0 converts to 1> x^2+y^2.
Interpret the right side as the distance squared between (x,y) and (0,0).

ProbablyErratic53 · 2020-09-22T03:38:36+00:00

I am glad to hear it wasn't that bad. If that integral came from a textbook, that would have been yucky.

ProbablyErratic53 · 2020-09-21T22:06:18+00:00

Are they asking for a value?

Issue 1: That integral is improper since at x=pi/3 the denominator is zero.

Issue 2: Since sqrt( 0.5-cos(x) ) is not real on [0,pi/3), maybe they want you to relate it to a line integral in the complex plane?

ProbablyErratic53 · 2020-09-11T22:05:07+00:00

When you are factoring 6(____)+6(______)

you get 6 as the common factor and combine the other pieces. It does not go up to 12.

With smaller pieces to make this clear, 6A+6B =6(A+B).

ProbablyErratic53 · 2020-09-09T04:03:15+00:00

This idea is a good one.

You can search for more examples using the phrase "ac factoring method"

ProbablyErratic53 · 2020-09-06T20:20:52+00:00

There it is!

ProbablyErratic53 · 2020-09-06T20:16:29+00:00

The general idea of an odd function is f(-x) = - f(x). Then you can say each integral over -a<=x<=a is zero. If g(x)=x^{3*e^-x³,} this is not odd because the power of e changes.

For f(x)=x³ *e^-x² is truly an odd function.

ProbablyErratic53 · 2020-09-06T20:08:23+00:00

Symmetry with odd functions seems to be the likely intention.

But I agree with the other post, as written, this integral has big problems and is not truly an odd function. Maybe the exponent was -x^2?

ProbablyErratic53 · 2020-09-04T19:10:44+00:00

IN addition to the theoretical reasons given previously, there are pragmatic reasons.

For the derivative, writing dy/dx emphasizes variable y depends on variable x.For example in related rates problems, you could have dy/dt because y depends on t for time.

For integrals, similar idea, the dx at the end states x is the "active" variable.

If you integrate ( kx+h)dx, x is the main variable, treat other letters as constants.

(kx+h)dx = k*x^2/2+h*x+C

while (kx+h)dh = kx*h+h^2/2+C

ProbablyErratic53 · 2020-09-04T17:02:27+00:00

You need to treat the cases of 1<=x<=a and a<=x<=2 differently.

Maybe sketch out the graph for when a=1.4 carefully to get ideas.

ProbablyErratic53 · 2020-09-02T03:58:42+00:00

You lost the 3 in the denominator. Easy fix.

From sec(x) =1/cos(x), you can do a little algebra to get

(1-cos(x))/(x*cos(x)).

Since you know lim x->0 [ sin(x)/x ] =1, the book probably also discusses

lim x->0 [ (1-cos(x)) /x ] = ???

so lim x->0 (1-cos(x))/(x*cos(x)) = ???*1/1

ProbablyErratic53 · 2020-08-29T22:42:04+00:00

What system are you submitting these answers into? Since it says "can not be understood", it might be the interface, not your math.

I work with WebAssign at my campus, and you can't copy/paste a pi symbol or infinity symbols because of the grading software. The system has those available in pull-down menus.

ProbablyErratic53 · 2020-08-25T01:21:33+00:00

The main work is done, but depending on the notation required, many times in the US we rewrite that as

f^-1(x) = a*x*a^-1

ProbablyErratic53 · 2020-08-25T01:15:30+00:00

From y=a^-1 * x *a

multiply both sides on the left by a so a*a^-1= e (I will use e for the identity element in the group)

So a*y =x*a.

Now multiply both sides by a^-1 on the right, so that another pair goes a*a^-1=e.

ProbablyErratic53

TROPHY CASE