Question

ib calc question:

http://i58.tinypic.com/121d5jm.png

Answer 1

@mathmale this is what I have so far:

Answer 2

http://i61.tinypic.com/2jg9pgo.jpg

Answer 3

You may want to start using snipping tool or something so we don't open links to sketchy sites to see your problem and see vulgar ad things that burn little girls minds. Like mine.

Answer 4

I think @jim_thompson5910 may be able to help you, he helps me with calculus all the time

Answer 5

sorry I didn't know there were ads; I have ad block. how do I use this snipping tool to put the picture in the post?

Answer 6

Well you snip the picture with the snipping tool, save it in your pictures and put it up by doing "Attach File"

Answer 7

I do not have ad block, I should look into how to get this you speak of because I'd rather not see male genitals and large female tooshies.

Answer 8

oh ok! yeah see adblock does what it does - blocks all these things so I forget they even exist! thanks for the tip. now does anyone know hot to solve this question?

Answer 9

Please for people who want to help him:

Click this and not the link. So much.. cleaner.

Answer 10

I won't be able to go into much detail on this problem tonight, as explained earlier.

I see you have an implicit function here, and expect that this relationship has a corresponding graph.

I'd suggest you find the derivative (dy/dx) using implicit differentiation. Why? Because this is the slope of the tangent line to the graph at any (x,y) that are in the domain/range.

Once you have that, find the NEGATIVE RECIPROCAL. This represents the SLOPE of the NORMAL LINE to the curve at the point of tangency.

Once you're at this point (c,ln c), you'll have the SLOPE of the normal line and the Y-INTERCEPT.

Experiment with this. I'll try to meet you again online tomorrow to continue this discussion.

Answer 11

@mathmale ok thank you so much! I will probably be on at the same time. I hope you saw my working?

Answer 12

This makes me scared for my future in Calc classes

Answer 13

Hmmm. Why do you think they give you a point of the form (c, ln(c))? Where did that ln(c) come from?

Answer 14

\[xe ^{-y} + e^y = 1 + x\]\[x + e^{2y} = e^y + xe^y\]\[ x(1-e^y)=x - xe^y = e^y - e^{2y} = (1-e^y)e^y\]=>\[x = e^y\]=>\[y = lnx \]

Answer 15

Perhaps, that may make things simpler.

Answer 16

\[\Large xe^{-y}+e^y=1+x\]
Let's differentiate.
\[\Large e^{-y}-xe^{-y} \frac{dy}{dx}+e^y \frac{dy}{dx}=1 => \frac{dy}{dx}(e^y-xe^{-y})=1-e^{-y}\]
\[\Large \frac{dy}{dx}=\frac{1-e^{-y}}{e^y-xe^{-y}}\]
This is the slope of tangent at a point say \(\LARGE (\alpha, \beta)\)
\[\Large \frac{dy}{dx}=\frac{1-e^{-\beta}}{e^\beta-\alpha e^{-\beta}}\]
Slope of normal..=>\[\Large \frac{dy}{dx}=\frac{\alpha e^{-\beta}-e^\beta}{1-e^{-\beta}}\]
At the point \[\LARGE (c,\ln c)\]
\[\Large \frac{dy}{dx}=\frac{c(1- c)}{c- 1}=-c\]
Equation of normal :
\[\LARGE y-\ln c = -c (x-c) =>y- \ln c=c^2 -cx\]
For y intercept..put x=0
\[\LARGE y- \ln c-c^2 =c^2+1 (given)\]
\[\LARGE 2c^2+\ln c=y-1\]
now we gotta solve for C..
did i make some calculation mistake?it seems tough moving ahead from here but I guess the method seems about right :O

Answer 17

Hey DLS. Instead of using point-slope, why don't we just use slope intercept? Makes more sense to me since we are given the y-intercept of the normal line. Using slope-intercept form, we have that \[\ln(c) = -(c)^2 + c^2 +1\] =>\[\ln(c) = 1\] => c = e

Answer 18

yes did't think of that XD