1. Well, I guess what i did is circular cone of radius R=1 and height R=1.

2. Originally Posted by Samantha
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

3. ## Can u help me with limits?

Find lim x->2 $\displaystyle (3x^2+5)$

4. Originally Posted by Samantha
Find lim x->2 $\displaystyle (3x^2+5)$
just plug in x = 2 in the formula and calculate the result, and that's it

5. lim x->3 $\displaystyle \frac {x-3} {Ix-3I}$

The limit does not exist, right?

6. Originally Posted by Samantha
lim x->3 $\displaystyle \frac {x-3} {Ix-3I}$

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

7. Originally Posted by Samantha
lim x->3 $\displaystyle \frac {x-3} {Ix-3I}$

The limit does not exist, right?
You can factor,
$\displaystyle \frac{(x-3)}{(x-3)I} = \frac{1}{I}$
So the limit is $\displaystyle \frac{1}{I}$ is $\displaystyle I\not = 0$.

8. Originally Posted by ThePerfectHacker
You can factor,
$\displaystyle \frac{(x-3)}{(x-3)I} = \frac{1}{I}$
So the limit is $\displaystyle \frac{1}{I}$ is $\displaystyle I\not = 0$.
i believe by $\displaystyle I$ she meant absolute values

9. Originally Posted by Jhevon
i believe by $\displaystyle I$ she meant absolute values
Yes, I meant absolute value

10. Originally Posted by Samantha
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

11. Originally Posted by Jhevon
I see what's the point of confusion here.

The volume of a right-cylinder is given by $\displaystyle V = \frac { \pi}{3} r^2 h$, however, in this particular cone, $\displaystyle h = r$. This is why Samantha made a distinction between the $\displaystyle r$'s and used $\displaystyle r \cdot r^2$. realizing that the first $\displaystyle r$ is for the height, she decided not to differentiate that one. However, this is incorrect. Janvdl is right. if $\displaystyle h = r$ it is perfectly valid to treat the $\displaystyle h$ as $\displaystyle r$ 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 $\displaystyle V = \frac { \pi}{3} r^2 h$ 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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