please help with problems from sample final D:

[michi]

I need help solving these problems because my ta is unavailable. But there are so many that I need help with that it seems overwhelming. Help on any of these would be greatly appreciated!

Please scroll down to see the problems as someone pointed out that I should not have the link.

The ones I am having trouble with are #2, 3, 5, 6 and 7.

#2 - I understand that I have to take the derivative of the equation, but then, I am stooped as how to proceed beyond that, and I'm not confident in taking the derivative of a square root, so can someone map this out for me?

#3 - I know this is an optimization problem, but I don't understand how to set it up and solve.

#5 I just don't know

#6 I really really don't know what to do with this one

#7 What are these asking for? Can someone please explain.

Thanks so much in advanced.

 

[happy]

You shouldn't post the direct link as many spambots can harvest your professor's email. You should just post copy and paste the pdf here instead of having us open a seperate file. Dr. Jian-Jun Xu even said, "When you send me emails, please include "math2a" in the subject. Otherwise, I may not be able to see it because of the large number of spam emails. "

 

[pka]

Prof Xu needs lessons in producing PDF files.
Your class should go together and give him a copy of Adobe Acrobat.

 

[Unco]

#2 -
\(\displaystyle \mbox{P(x) = x\sqrt{x^2 + 1}}\)

I understand that I have to take the derivative of the equation, but then, I am stooped as how to proceed beyond that, and I'm not confident in taking the derivative of a square root, so can someone map this out for me?

To differentiate \(\displaystyle \mbox{\sqrt{x^2 + 1}}\) with respect to \(\displaystyle \mbox{x}\),
write as \(\displaystyle \mbox{(x^2 + 1)^{\frac{1}{2}}}\) and apply the chain rule:

\(\displaystyle \mbox{\frac{d}{dx} (x^2 + 1)^{\frac{1}{2}} = \frac{1}{2}(x^2 + 1)^{-\frac{1}{2}} \cdot 2x = x(x^2 + 1)^{-\frac{1}{2}}}\)

Are you ok to apply the product rule to differentiate \(\displaystyle \mbox{P(x)}\) now?

I'm sure you must have covered the content #2 involves in class; which parts specifically are troubling you?

 

[stapel]

For reference:

. . . . .2) Let \(\displaystyle \large{p(x)\,=\,x\,\sqrt{x^2\,+\,1}}\)
. . . . .a) Find the intervals on which p(x) is increasing
. . . .. . . or decreasing.
. . . . .b) Find the local max/min values.
. . . . .c) Find the intervals of concavity and the inflection
. . . .. . . points.
. . . . .d) Sketch the graph, based on the above information.


. . . . .3) If 1200 cm² of material is available
. . . . . . .to make a box with a square base and an open top,
. . . . . . .find the largest possible volume of the box.


. . . . .5) Express the following limit as a definite integral,
. . . . . . .where delta-x = (2pi)/n, x_i = (i)(delta-x),
. . . . . . .and i = 0, 1, ..., n. Use the Fundamental Theorem of
. . . . . . .Calculus to evaluate the integral.

. . . . . . . .\(\displaystyle \large{\begin{array}{c}lim\\n\rightarrow \infty\end{array}\,\sum^n_{n=1}\,\left(x_i\,+\,{x_i}^2\right)\,\Delta x}\)


. . . . .6) Use the Mean Value Theorem to prove the following:

. . . . . . .|cos(x) - cos(y)| < |x - y|


. . . . .7-a) Find the following limit:

. . . . . . . .\(\displaystyle \large{\begin{array}{c}lim\\x\rightarrow 1\end{array}\,\frac{\tan{(4)(x\,-\,1)}}{(x\,-\,1)}\)

. . . . .7-b) Differentiate y = sqrt[sec(x³)]

Note to tutors: It is not clear, in the original, whether the "x - 1" in the numerator in (7-a) is meant to be inside the tangent, along with the "4".

Eliz.

 

[Daniel_Feldman]

For reference:

. . . . .2) Let \(\displaystyle \large{p(x)\,=\,x\,\sqrt{x^2\,+\,1}}\)
. . . . .a) Find the intervals on which p(x) is increasing
. . . .. . . or decreasing.
. . . . .b) Find the local max/min values.
. . . . .c) Find the intervals of concavity and the inflection
. . . .. . . points.
. . . . .d) Sketch the graph, based on the above information.


. . . . .3) If 1200 cm² of material is available
. . . . . . .to make a box with a square base and an open top,
. . . . . . .find the largest possible volume of the box.


. . . . .5) Express the following limit as a definite integral,
. . . . . . .where delta-x = (2pi)/n, x_i = (i)(delta-x),
. . . . . . .and i = 0, 1, ..., n. Use the Fundamental Theorem of
. . . . . . .Calculus to evaluate the integral.

. . . . . . . .\(\displaystyle \large{\begin{array}{c}lim\\n\rightarrow \infty\end{array}\,\sum^n_{n=1}\,\left(x_i\,+\,{x_i}^2\right)\,\Delta x}\)


. . . . .6) Use the Mean Value Theorem to prove the following:

. . . . . . .|cos(x) - cos(y)| < |x - y|


. . . . .7-a) Find the following limit:

. . . . . . . .\(\displaystyle \large{\begin{array}{c}lim\\x\rightarrow 1\end{array}\,\frac{\tan{(4)(x\,-\,1)}}{(x\,-\,1)}\)

. . . . .7-b) Differentiate y = sqrt[sec(x³)]

Note to tutors: It is not clear, in the original, whether the "x - 1" in the numerator in (7-a) is meant to be inside the tangent, along with the "4".

Eliz.

Thanks Eliz.

For 7(a), I'm assuming the numerator is tan4(x-1)=tan(4x-4). I may be wrong.

If I'm right however, here we go.

Substituting 1 yields the indeterminate form 0/0 so we use L'Hopital's rule.

Taking the derivative of the numerator and the denominator yields


[4sec^2(4x-4)]/1=4sec^2(4x-4)

Substituting 1 now yields our answer

4Sec^2(4-4)=4sec^2(0)=4(1)=4

 

[galactus]

For reference:

. . . . .2) Let \(\displaystyle \large{p(x)\,=\,x\,\sqrt{x^2\,+\,1}}\)
. . . . .a) Find the intervals on which p(x) is increasing
. . . .. . . or decreasing.
. . . . .b) Find the local max/min values.
. . . . .c) Find the intervals of concavity and the inflection
. . . .. . . points.
. . . . .d) Sketch the graph, based on the above information.


. . . . .3) If 1200 cm² of material is available
. . . . . . .to make a box with a square base and an open top,
. . . . . . .find the largest possible volume of the box.

Write an equation expressing the surface area of the box:

The base is square, so its area is \(\displaystyle x^{2}\). See?. The sides, there

are 4, have length x and height y. OK?. So, we have\(\displaystyle x^{2}+4xy=1200\)

The volume is just \(\displaystyle V=x^{2}y\).

Now, solve the surface area equation for x or y, whichever you

choose; sub it into the volume equation, differentiate, set to 0 and solve.


. . . . .5) Express the following limit as a definite integral,
. . . . . . .where delta-x = (2pi)/n, x_i = (i)(delta-x),
. . . . . . .and i = 0, 1, ..., n. Use the Fundamental Theorem of
. . . . . . .Calculus to evaluate the integral.

. . . . . . . .\(\displaystyle \large{\begin{array}{c}lim\\n\rightarrow \infty\end{array}\,\sum^n_{n=1}\,\left(x_i\,+\,{x_i}^2\right)\,\Delta x}\)


. . . . .6) Use the Mean Value Theorem to prove the following:

. . . . . . .|cos(x) - cos(y)| < |x - y|


. . . . .7-a) Find the following limit:

. . . . . . . .\(\displaystyle \large{\begin{array}{c}lim\\x\rightarrow 1\end{array}\,\frac{\tan{(4)(x\,-\,1)}}{(x\,-\,1)}\)

. . . . .7-b) Differentiate y = sqrt[sec(x³)]

Note to tutors: It is not clear, in the original, whether the "x - 1" in the numerator in (7-a) is meant to be inside the tangent, along with the "4".

Eliz.

 

[Unco]

Use the Mean Value Theorem to prove the following:

\(\displaystyle \mbox{|\cos{(x)} - \cos{(y)}| \leq |x - y|}\)

My notation is rusty but it's pretty clear you want to use Mean Value Theorem definition and the fact that \(\displaystyle \mbox{ \left(\cos{(u)}\right)' = -\sin{u} \Rightarrow \left|\left (\cos{(u)}\right)'\right| \leq 1}\).