Archive: 26 posts

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I need help with my homework >.< I have no idea how to work this question out Equation of the tangent to y= -2x^2 at x=1 I first tried substituting x into the equation... but it = -2.... which i don't think is right.... I'm not asking for an answer, if someone can help me how to work it out, i would be for ever greatful!

2009-09-15 12:06:00 Author: snowyjoe Posts: 509

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You need to replace the x with 1, no? So it's y=(-2 times 1)^2=-2^2 Edit: My calculator tells me something different.... Maybe I'm doing something wrong. It's been too long.

2009-09-15 12:27:00 Author: Syroc Posts: 3193

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Won't it be 1 to the power of 2 first? Anyway, I've acturally already tried substituting/replacing x (what ever you want to call it) and I don't think it's the correct answer.

2009-09-15 12:40:00 Author: snowyjoe Posts: 509

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Well, that wouldn't matter because 1^2 is 1. But you don't do that anyway. For what it's worth my calculator says -2 as well. You are searching for y at the position of x=1, right?

2009-09-15 12:42:00 Author: Syroc Posts: 3193

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I got it! Not only do i replace x but I also have to find the Derivative of the equation as well! Thats why my answer isn't correct.... But how do you find the dy/dx? :'(

2009-09-15 12:50:00 Author: snowyjoe Posts: 509

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Differentiation: Multiply by the power so X^2 = 2X and then -1 from the power so it's : dy/dx = X^2 = 2X^2-1 = 2X^1 = 2X EDIT: So for your question you find the value of Y by subbing in X. You then use DY/DX to find M (the gradient). To find the forumla you then use the equation: y - y' M = ----- x - x' Y' and x' are the values you have found of y and x. Then M is the gradient. Then you'll get something like this: y - 3 5 = ------ x - 4 multiply it out so it is y-3 = 5(X-4) y -3 = 5X -20 y= 5X -17

2009-09-15 12:52:00 Author: Shermzor Posts: 1330

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I feeling quite stupid at the moment. Only a few years ago I could solve these things without a problem. This might not motivate those that are still in school, but the chances that you will need calculus in your later life are very slim. You're much more likely to need statistics.

2009-09-15 12:55:00 Author: Syroc Posts: 3193

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I feeling quite stupid at the moment. Only a few years ago I could solve these things without a problem. This might not motivate those that are still in school, but the chances that you will need calculus in your later life are very slim. You're much more likely to need statistics. That's why the current education system sucks! Alot!!! Anyway thanks for the very helpful info Shermzor! You're a life saver!!!!!

2009-09-15 12:58:00 Author: snowyjoe Posts: 509

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That's why the current education system sucks! Alot!!! Anyway thanks for the very helpful info Shermzor! You're a life saver!!!!! I'm sitting in my study room at lunch time with nothing to do and a fat folder of maths next to me so no problem Chances are you'll be doing intergration after that so if you want brownie points you can read up on that but it's basically the reverse of differentiation

2009-09-15 13:02:00 Author: Shermzor Posts: 1330

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Read!? But text books make it sound so hard, boring and confusing >.< Then you have teachers that just read out of the text book.... which is even more unhelpful! I think drawing LBP contraptions in my math book would benefit me better than a brownie

2009-09-15 13:07:00 Author: snowyjoe Posts: 509

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This might not motivate those that are still in school, but the chances that you will need calculus in your later life are very slim. You're much more likely to need statistics. That's a bit of a silly thing to say. The majority of people won't need 95% of what they learn in school when they get to later life. It doesn't mean it's not worth knowing. The education system is designed to give you a firm foundation in a broad range of subjects, leading to specialism later on. Also, no-one with any sense does stats longhand, you'd always use a computer for that. Calculus is often faster by hand, so which one is more useful to learn properly? Assuming that you are not likely to need either

2009-09-15 13:18:00 Author: rtm223 Posts: 6497

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Yeah true, but honestly they should put much more focus on statistics than they are doing now. Most schools have computers by now I assume. Also learning SPSS in school would be horrible. No matter how useful the program is.

2009-09-15 13:23:00 Author: Syroc Posts: 3193

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Oh darn! This was right up my alley, but I was beaten to it. ...actually, I'm confused by your help there, Shermzor... did you intentionally use different equations and points? For your specific problem, for which y = -2x^2, dy/dx will be -4x, using the chain rule. This tells us the slope of the line tangential to y = -2x^2 at any value of x along the curve. We are concerned only with the line tangential to the curve at x = 1, so the slope of the line is equal to -4(1) = -4. Now, we must use the point-slope equation (or whatever y'all call it where you are) to find a specific equation for the line. As Sherm wrote, we need to us this (slope = M): y - y' M = ----- x - x' To find x' and y', simply plug x = 1 into the original equation to obtain: y = -2(1)^2 = -2. Clearly, x' = 1, and we just found y' to be -2. Now, we have: y - (-2) -4 = -------- ==> y + 2 = -4(x - 1) ==> y = -4x + 2 x - 1

2009-09-15 13:24:00 Author: comphermc Posts: 5338

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Oooh, the memories are coming back. ^.^

2009-09-15 13:26:00 Author: Syroc Posts: 3193

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Yer I pulled the numbers out of the air I'm not touching a calculator in my breaktimes!

2009-09-15 15:32:00 Author: Shermzor Posts: 1330

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Whatever the answer was.....I agree lol!!!!!

2009-09-15 15:41:00 Author: mrsvista Posts: 755

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thread hijack. http://i26.tinypic.com/2yoq3h5.jpg I would appreciate if someone could help me, be it only one answer, I can use every single thing

2009-09-15 15:44:00 Author: oldage Posts: 2824

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To be honest seems pretty simple just watch out for minus times a minus equals a plus and silly mistakes like that. Also with your surds you might want to simplify them out at the end. EDIT: aha I can open the image I might have a bop at these ask me if you get stuck on one

2009-09-15 15:55:00 Author: Shermzor Posts: 1330

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What does the ":" mean in these problems? Do they want a ratio? Edit: OHH!! Divide! Got it. Same thing as ratio, but I've never seen it that way before.

2009-09-15 16:22:00 Author: comphermc Posts: 5338

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What does the ":" mean in these problems? Do they want a ratio? Edit: OHH!! Divide! Got it. Same thing as ratio, but I've never seen it that way before. rofl, that's how we write "divide" here! @sherm: I'm having trouble with 16)....

2009-09-15 16:34:00 Author: oldage Posts: 2824

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For number 16, I won't give you the answer, but write the left hand side in the numerator (above the fraction) and the right hand side in the denominator (under the fraction). Anything that appears in both the top and bottom can be cancelled. Just another little hint: 2/4 = 1/2. This may be obvious, but you never know.

2009-09-15 16:53:00 Author: comphermc Posts: 5338

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http://farm1.static.flickr.com/104/297016041_06159f87e8_o.jpg (more (http://www.darkroastedblend.com/2006/11/new-breakthroughs-in-mathematics.html))

2009-09-15 16:57:00 Author: Syroc Posts: 3193

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For number 16, I won't give you the answer, but write the left hand side in the numerator (above the fraction) and the right hand side in the denominator (under the fraction). Anything that appears in both the top and bottom can be cancelled. Just another little hint: 2/4 = 1/2. This may be obvious, but you never know. ooh I get it, thanks for the hint!

2009-09-15 17:04:00 Author: oldage Posts: 2824

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Sorry got picked up early so I'm home now Though I'll be busy trying to find out where the hell I enter my batman arkham aylum code...

2009-09-15 17:26:00 Author: Shermzor Posts: 1330

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http://www.guzer.com/pictures/findx.jpg

2009-09-16 06:22:00 Author: snowyjoe Posts: 509

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Meh. i was just about to ask a question about my homework too. i really don't want to be a leach, but its too late to ask someone for help in real life, and my trig book doesn't give any explanation as to what i'm supposed to do. (i'm sorry, i just don't want to get a 0) mine's just a simple proof in trig 101. shouldn't be to hard compared to the calculus question... "show that each of the following statements is an identity by transforming the left side of each one into the right side" tan?Θ + 1 = sec?Θ secΘ cotΘ - sinΘ = (cos?Θ/(sinΘ (sinΘ - cosΘ? - 1 = -2 sinΘ cosΘ I already know my reciprocal and quotient trigonometric identities - i just don't see how to apply them with the addition and subtraction included in the questions. i don't blame anyone who calls me a leech for this, but the only other place i could possibly ask this late at night is at the other community forum i go to - and knowing them they would probably answer the question in assembly language or something just to spite me. sorry ;_;

2009-09-16 07:19:00 Author: brb_gymnastics Posts: 32

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