i've been sick....

[jadorexxx]

I've been sick a whole week, and while it may seem like my teacher might be the one to ask, you are wrong. lol.

Analyze, find any intercepts, relative extrema, points of inflection, and asymptotes, and sketch the graph of the function

Maximum Area: Find the dimensions of the rectangle of maximum area, with sides parallel to the coordinate axes, that can be inscribed in the ellipse given by

 

[stapel]

"Intercepts" and "asymptotes" were covered back in algebra, so you should be able to do them. And I doubt that first derivatives, max/min points, second derivatives, concavity, and points of inflection were covered in just a week, so you should at least be able to do the derivatives.

Please reply showing what you have done, and clearly stating where you are stuck.

Please note that we cannot teach lessons here, so if the problem is that "I was out all week and don't know this stuff at all", we would need to send you to lessons sites where you can learn the content, before you could attempt the exercises. Only then could we usefully provide advice, if you have really never seen this stuff at all.

Thank you.

Eliz.

 

[galactus]

Let's show a general case for the ellipse problem. Then all you have to do is plug in your a and b. Okey-doke?.

The area of the rectangl is \(\displaystyle (2x)(2y)\)

The entire thing will then have area \(\displaystyle A=4xy\)

The equation of an ellipse is \(\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Solve for y: \(\displaystyle \frac{b}{a}\sqrt{a^{2}-x^{2}}\)

Insert into Area equation:

\(\displaystyle A(x)=\frac{4bx}{a}\sqrt{a^{2}-x^{2}}\).

Now differentiate:

\(\displaystyle A'(x)=\frac{4b}{a}((x)(\frac{-x}{\sqrt{a^{2}-x^{2}}})+(sqrt{a^{2}-x^{2}})(1))=\frac{4b(a^{2}-2x^{2})}{a\sqrt{a^{2}-x^{2}}\)

Skipping the algebra steps(you can do that part), set to 0 and solve we find that A'(x)=0 when \(\displaystyle x=\frac{a}{\sqrt{2}}\)

Inserting this into A(x) we find that max area is \(\displaystyle 2ab\)

 

[jadorexxx]

for the 1st question...

I changed it to

I got the 1st derivative to -20x/(x^2-9)^2... and it looks wrong? I don't know.

I then tried and changed the problem to this...
y = 1 + 10/(x^2-9)

And I know that 10/(x^2-9) part approaches infinity so an asymptote is at 1?



????

 

[stapel]

for the 1st question... I got the 1st derivative to -20x/(x^2-9)^2... and it looks wrong?

Why? What do you mean by "it looks wrong" (other than that the "dy/dx =" part has been omitted)? Please clarify.

I then tried and changed the problem to...y = 1 + 10/(x^2-9) And I know that 10/(x^2-9) part approaches infinity so an asymptote is at 1?

If you are referring to the horizontal asymptote then, yes, as the rule you learned back in algebra taught you (see second lesson link below), the horizontal asymptote is y = 1.

. . . . .FreeMathHelp lesson: Finding Vertical Asymptotes

. . . . .FreeMathHelp lesson: Finding Horizontal Asymptotes

. . . . .FreeMathHelp lesson: Finding Zeroes (x-intercepts)

. . . . .FreeMathHelp lesson: Finding Critical Points

When you reply, please show how far you've gotten on finding the vertical asymptotes (see first lesson link above), the x-intercepts (see third lesson link above), the critical points you found for dy/dx (see fourth lesson link above), and how far you've gotten in applying the Quotient Rule again (to dy/dx) to find d²y/dx².

Thank you.

Eliz.