helpp please due soon | Mathematics | Page 1

pulsar lanceBOT Nov 21, 2024, 3:37 AM

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green kelp Nov 21, 2024, 3:51 AM

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!status

pulsar lanceBOT Nov 21, 2024, 3:51 AM

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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

bold axle Nov 21, 2024, 3:54 AM

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idk how to solve it

pulsar lanceBOT Nov 21, 2024, 5:27 AM

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helpp please due soon

winged stirrup Nov 21, 2024, 6:32 AM

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bold axle Nov 21, 2024, 6:16 PM

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Whaat I'm sorry

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What's that 😭

winged stirrup Nov 21, 2024, 7:44 PM

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bold axle What's that 😭

You want to find the area under the curve enclosed by the two functions between an interval. If you want to find the integral over that interval, what bounds would you use?

Also, if you find the area under the curve for both individually, how can they 1. Be written together and 2. Account for the fact that you will have an overlapping area

bold axle Nov 21, 2024, 7:45 PM

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Isn't c supposed to be the ans

winged stirrup Nov 21, 2024, 7:46 PM

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bold axle Isn't c supposed to be the ans

Why do you think so?

bold axle Nov 21, 2024, 7:46 PM

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rn_image_picker_lib_temp_43f651cb-0f27-4ec6-9509-e3c2c06b1eef.jpg

winged stirrup Nov 21, 2024, 7:47 PM

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I’ll give you a hint, -4 to 5 is not the correct bounds

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We don’t want to find the ENTIRE area under the curve, because that would include space we’re not counting (namely, the area under x^2 which is NOT between the line and the x^2 curve)

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Here’s a visual representation of your interval and the two curves

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If we want the area enclosed by x+12 and x^2, what bounds would we use?

bold axle Nov 21, 2024, 7:57 PM

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May ik why -4 and 5 aren't the correct bounds
When I tried solving the qs someone solved it like the way I did

winged stirrup Nov 21, 2024, 7:58 PM

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bold axle May ik why -4 and 5 aren't the correct bounds When I tried solving the qs some...

Look at the graph. Between -4 and 5, is the entire area enclosed by x^2 and x+12?

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Posting without the bounds to make it even more explicit

bold axle Nov 21, 2024, 7:58 PM

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winged stirrup Look at the graph. Between -4 and 5, is the entire area enclosed by x^2 and x+12...

No

winged stirrup Nov 21, 2024, 7:59 PM

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bold axle No

So, between what values is there an enclosure?

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Also, is (x+12)-x^2 equal to x^2-x-12 like you said above?

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That should be another indication of why answer choice C is incorrect (regardless of bounds)

bold axle Nov 21, 2024, 8:02 PM

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So is it x^2-x-12 and -x^2+x+x

winged stirrup Nov 21, 2024, 8:02 PM

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bold axle So is it x^2-x-12 and -x^2+x+x

Well, there’s no and

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It’s one or the other

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Take away the parentheses since addition is associative, x+12+(-x^2), we also know that addition is commutative, so how can we rephrase this expression with -x^2 at the front?

winged stirrup Nov 21, 2024, 8:04 PM

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bold axle So is it x^2-x-12 and -x^2+x+x

I think you meant to put a 12 on the end here too instead of the second x

bold axle Nov 21, 2024, 8:05 PM

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Oh yeah

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Is it either d or g

winged stirrup Nov 21, 2024, 8:18 PM

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bold axle Is it either d or g

Also consider there’s another area that IS between -4 and 5 which is enclosed

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So we are going to need the other equation you posted earlier, and the intuition for that is that it’s enclosed in the other direction. The line sets the bounds within the x^2 curve, but also an area outside of it

winged stirrup Nov 21, 2024, 8:21 PM

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winged stirrup Here’s a visual representation of your interval and the two curves

bold axle Nov 21, 2024, 8:45 PM

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It's d right

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😀

winged stirrup Nov 21, 2024, 8:47 PM

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bold axle It's d right

I believe so!

bold axle Nov 21, 2024, 8:47 PM

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I'm on my last attempt should I try 👀

winged stirrup Nov 21, 2024, 8:47 PM

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Send it

winged stirrup Nov 21, 2024, 8:52 PM

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bold axle I'm on my last attempt should I try 👀

Were we right?

bold axle Nov 21, 2024, 9:04 PM

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Yeah

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Let's go

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Thank u so much

winged stirrup Nov 21, 2024, 9:11 PM

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bold axle Thank u so much

Np

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Been years since I’ve done calc so never 100% certain if i’m right lmao

bold axle Nov 21, 2024, 9:12 PM

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Can u help me solve another qs

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I hate calc

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I'm a science student not math

winged stirrup Nov 21, 2024, 9:12 PM

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bold axle Can u help me solve another qs

Of course, send em

bold axle Nov 21, 2024, 9:13 PM

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winged stirrup Nov 21, 2024, 9:21 PM

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bold axle

Do you know how to find the integral without rotating first off

bold axle Nov 21, 2024, 9:26 PM

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That's a very good question

winged stirrup Nov 21, 2024, 9:26 PM

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bold axle That's a very good question

Think about it like we did before

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What are the bounds here?

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Graph so you have a visual

bold axle Nov 21, 2024, 9:28 PM

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U live in canada?

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Why is the graph looking like that

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I'm so so sorry but what's that 😭😭😭

winged stirrup Nov 21, 2024, 9:30 PM

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bold axle I'm so so sorry but what's that 😭😭😭

It’s the curve arctan of x, y=1/6, and shows the bound at 7 (imagine a bound at 0, when I put it in it’ll look too murky so just imagine)

winged stirrup Nov 21, 2024, 9:31 PM

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bold axle U live in canada?

no

bold axle Nov 21, 2024, 9:33 PM

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winged stirrup It’s the curve arctan of x, y=1/6, and shows the bound at 7 (imagine a bound at ...

Okay what should I do next

bold axle Nov 21, 2024, 9:33 PM

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winged stirrup no

Oh we have same timings

winged stirrup Nov 21, 2024, 9:33 PM

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bold axle Okay what should I do next

Think about what the bounds will be

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What regions are enclosed and over what intervals?

bold axle Nov 21, 2024, 9:35 PM

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0 and 7

winged stirrup Nov 21, 2024, 9:36 PM

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bold axle 0 and 7

Give the interval corresponding to the enclosed regions

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My hint is that there are 2

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But we’ll start on 0 and end on 7, yes

bold axle Nov 21, 2024, 9:37 PM

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Like do u mean 2pi

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So it may be either e or g

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We are talking abt part a right

winged stirrup Nov 21, 2024, 9:43 PM

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bold axle We are talking abt part a right

Yes

winged stirrup Nov 21, 2024, 9:43 PM

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bold axle Like do u mean 2pi

No

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I mean what are the enclosed regions

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Find the bounds before you do anything else

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It’ll knock off half of the answer choices

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and then selecting the right equation will knock off another half from the remaining

bold axle Nov 21, 2024, 9:44 PM

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1/6

winged stirrup Nov 21, 2024, 9:45 PM

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bold axle 1/6

that’ll be the y value (which we know, because y=1/6 will intersect it at y=1/6), but what would the x value be (as the x is what defines our bounds)

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Stated in a leading way, If arctan(x)=1/6, how do we find x?

Hint: what is the inverse function of arctan?

bold axle Nov 21, 2024, 9:49 PM

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1/x^2+1

winged stirrup Nov 21, 2024, 9:49 PM

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bold axle 1/x^2+1

What is the inverse function of arctan?

bold axle Nov 21, 2024, 9:50 PM

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Isn't that supposed to be the inverse?

winged stirrup Nov 21, 2024, 9:51 PM

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bold axle Isn't that supposed to be the inverse?

No

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What is the inverse function of arctan?

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Arctan is the inverse of tan, and BLANK is the inverse of arctan

bold axle Nov 21, 2024, 9:52 PM

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Tan?

winged stirrup Nov 21, 2024, 9:52 PM

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Yes

bold axle Nov 21, 2024, 9:52 PM

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We don't need the deriv?

winged stirrup Nov 21, 2024, 9:52 PM

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Nope

bold axle Nov 21, 2024, 9:52 PM

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Ohh

winged stirrup Nov 21, 2024, 9:52 PM

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So if arctan(x)=1/6, x=?

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Don’t compute the actual x, just keep it in function form (as this is what the problem does in the solutions, and the answer is a crazy decimal)

bold axle Nov 21, 2024, 9:54 PM

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0.165149

winged stirrup Nov 21, 2024, 9:55 PM

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bold axle 0.165149

Keep it in function form, the decimal isn’t used in the problem

bold axle Nov 21, 2024, 9:56 PM

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Tan 1/6

winged stirrup Nov 21, 2024, 9:57 PM

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bold axle Tan 1/6

Yes

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So what are our bounds

bold axle Nov 21, 2024, 9:57 PM

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0, tan1/6, 7

winged stirrup Nov 21, 2024, 9:57 PM

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Yes!

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Ok, next determine the equations that’ll give you the area of the enclosed regions (don’t worry about rotating just yet, that’ll be our last step)

bold axle Nov 21, 2024, 9:59 PM

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Arctanx^2 and 1/6^2

winged stirrup Nov 21, 2024, 9:59 PM

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Don’t worry about rotating yet

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From 0 to tan(1/6) what do we need to do to ONLY get the area of the enclosed region?

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Look at the graph again if need be

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Just like the last problem we did

winged stirrup Nov 21, 2024, 10:02 PM

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winged stirrup Graph so you have a visual

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Extend that to x=7 I had to crop so the first enclosed space is actually visible

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If you’re not allowed to look at graphs, the easiest way to know which function is higher than the other at a particular bound is to just plug in an x in the bound and see which value is greater (for us x=0 to start with). For us, we know that y=1/6 is above at first because arctan(0)=0

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and then we can check again in the next bound

bold axle Nov 21, 2024, 10:05 PM

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We won't have graphs in our test

winged stirrup Nov 21, 2024, 10:05 PM

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Yes so follow what I said

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You can also find the overlapping points that show your separated bounds by setting the functions equal to each other (because they overlap when they’re equal in x and y value)

bold axle Nov 21, 2024, 10:07 PM

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So it's h

winged stirrup Nov 21, 2024, 10:07 PM

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Wait wait wait

bold axle Nov 21, 2024, 10:07 PM

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🥸

winged stirrup Nov 21, 2024, 10:08 PM

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It’s not, but I also said don’t rotate yet

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What’s the equation (with integral 0 to tan(1/6)) that’ll get you the enclosed area, knowing y=1/6 is on top, and that what’s below arctanx (because it’s the lower function at that point) isn’t in the enclosed space

bold axle Nov 21, 2024, 10:11 PM

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Acrtan 1/6 will be in the middle like top in the first and below in the second

winged stirrup Nov 21, 2024, 10:12 PM

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Yes, now give the equation for the first interval (don’t write out the integral sign and stuff just give me the plain algebraic expression)

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We want everything below y=1/6 WITHOUT what’s below arctan(x), as that’s our enclosed space. What’s the equation

bold axle Nov 21, 2024, 10:12 PM

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1/6-arctanx

winged stirrup Nov 21, 2024, 10:12 PM

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Beautiful

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and for the second?

bold axle Nov 21, 2024, 10:13 PM

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Arctanx-1/6

winged stirrup Nov 21, 2024, 10:13 PM

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Awesome

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Now we want to know what the integral will look like if it’s a circle

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Because a 360 revolution is a circle

bold axle Nov 21, 2024, 10:14 PM

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We double it bcz we have a bigger and a smaller circle

winged stirrup Nov 21, 2024, 10:15 PM

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No

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If you want to find the area of a circle, what equation do you use

bold axle Nov 21, 2024, 10:15 PM

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Pir2

winged stirrup Nov 21, 2024, 10:16 PM

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Ok

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What’s our radius

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Hint: You’ve already figured it out

bold axle Nov 21, 2024, 10:19 PM

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Arctan1/6

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Or tan1/6

winged stirrup Nov 21, 2024, 10:20 PM

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No

winged stirrup Nov 21, 2024, 10:23 PM

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bold axle Arctan1/6

Think hard about it. We’re not going to have a constant value as the radius, because the circle changes size across. What would represent the value of the thing being rotated around the x axis, let’s say for the first interval

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Because when it’s rotated (remembering that the radius is HALF of the circle), and our enclosed space is the thing being rotated (and it constitutes half of the circle), what’s our radius?

bold axle Nov 21, 2024, 10:25 PM

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7

winged stirrup Nov 21, 2024, 10:28 PM

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bold axle 7

I’ll make a graph to explain

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The distance between the y values of the two points represents the radius at a particular value x=4. If we JUST use arctan(4)=y, we’ll get the distance from 4 on the x axis to the top function. But we only want the enclosed space, so what can we get rid of to make the radius ONLY the enclosed space?

winged stirrup Nov 21, 2024, 10:31 PM

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winged stirrup The distance between the y values of the two points represents the radius at a p...

bold axle Nov 21, 2024, 10:33 PM

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Ohhh okay

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I get it

winged stirrup Nov 21, 2024, 10:34 PM

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So what’s the radius IN GENERAL (not for x=4, the generalization is just x because we want it at any real x between 0 and 7)

bold axle Nov 21, 2024, 10:34 PM

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Is the ans d? For part a

winged stirrup Nov 21, 2024, 10:35 PM

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bold axle Is the ans d? For part a

Yes

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The radius is just the expressions we found before

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arctanx-1/6 and 1/6-arctanx

bold axle Nov 21, 2024, 10:38 PM

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How abt part b

winged stirrup Nov 21, 2024, 10:41 PM

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bold axle How abt part b

Ok, first we want to find the inverses of the functions

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Because we want y to be equal to the x value of the function, leading from the fact that we’re now rotating around the y axis

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So find a way to get x on its own in each function

bold axle Nov 21, 2024, 10:54 PM

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ok

winged stirrup Nov 21, 2024, 10:54 PM

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So it’ll be x in terms of y

bold axle Nov 21, 2024, 11:38 PM

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is it a?

winged stirrup Nov 21, 2024, 11:48 PM

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bold axle is it a?

No

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Don’t guess, do the work

bold axle Nov 21, 2024, 11:51 PM

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🧍‍♀️🧍‍♀️

winged stirrup Nov 22, 2024, 12:13 AM

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bold axle 🧍‍♀️🧍‍♀️

Did you do the function inversions

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y=arctan(x). What is x equal to?
x=0. Already know what x equals
x=7. Already know what x equals
y=1/6. Already know what y equals

bold axle Nov 22, 2024, 4:54 AM

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Tan(1/6

bold axle Nov 24, 2024, 8:11 AM

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.close

#helpp please due soon