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Math Help - Another question

  1. #16
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    Well, I guess what i did is circular cone of radius R=1 and height R=1.
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  2. #17
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Samantha View Post
    Well, I guess what i did is circular cone of radius R=1 and height R=1.
    why would you do that if the question says "when radius equals 2 feet?" and why didn't you differentiate?

    janvdl's method was right, look it over and make sure you understand what he did, if not, ask specific questions
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  3. #18
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    Can u help me with limits?

    Find lim x->2 (3x^2+5)
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  4. #19
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Samantha View Post
    Find lim x->2 (3x^2+5)
    just plug in x = 2 in the formula and calculate the result, and that's it
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  5. #20
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    lim x->3 \frac {x-3} {Ix-3I}


    The limit does not exist, right?
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  6. #21
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Samantha View Post
    lim x->3 \frac {x-3} {Ix-3I}


    The limit does not exist, right?
    correct, it approaches -1 if we approach from the left of 3, but +1 if we approach from the right of 3. i hope you know why
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  7. #22
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    Quote Originally Posted by Samantha View Post
    lim x->3 \frac {x-3} {Ix-3I}


    The limit does not exist, right?
    You can factor,
    \frac{(x-3)}{(x-3)I} = \frac{1}{I}
    So the limit is \frac{1}{I} is I\not = 0.
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  8. #23
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    You can factor,
    \frac{(x-3)}{(x-3)I} = \frac{1}{I}
    So the limit is \frac{1}{I} is I\not = 0.
    i believe by I she meant absolute values
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  9. #24
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    Quote Originally Posted by Jhevon View Post
    i believe by I she meant absolute values
    Yes, I meant absolute value
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  10. #25
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Samantha View Post
    Yes, I meant absolute value
    ok, so go with my answer, TPH mistook I for a variable "I" and not absolute values. Or i shouldn't say mistook, it is I! there is an a symbol on your keyboard that you can use for absolute values. on US keyboards it is the same key as "\", just hold down "shift" while you're pressing that key and you get "|", which is the absolute value sign
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  11. #26
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    Quote Originally Posted by Jhevon View Post
    I see what's the point of confusion here.

    The volume of a right-cylinder is given by V = \frac { \pi}{3} r^2 h, however, in this particular cone, h = r. This is why Samantha made a distinction between the r's and used r \cdot r^2. realizing that the first r is for the height, she decided not to differentiate that one. However, this is incorrect. Janvdl is right. if h = r it is perfectly valid to treat the h as r in every way. in fact, this is a prevalent related rates technique. a lot of questions like this require you to solve for one variable in terms of another so you can differentiate a single variable function, we do not make distinctions between the same variable despite the fact it came from two different quantities.

    even if your method was right Samantha, you would still get it wrong, why did you plug in 1 for the r that represents the height? if h is equal to r, then when r = 2, the height would be the same. and looking at it again, it seems you didn't differentiate at all. i'm sorry but what you did is confusing (i am looking at your last attempt at the question). can you explain exactly what you were doing so that we can clear up any misconceptions you have?
    Yes, i did indeed use V = \frac { \pi}{3} r^2 h and i tried using implicit differentiation(like you taught me Jhevon!)

    But then Samantha gave me this formula and i realised that h = r.
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