# Math Help - So I'm having some trouble...

1. ## So I'm having some trouble...

On a good number of questions. I realize I'm new here, but I hope to stick around. I have left a more sizable message after the problems, in case you're interested.

Here goes:

5. "The base of a solid object is the region in the xy-plane bounded by the graphs of y=x^2 and y=4. Every verticle cross section parallel to the y-axis is a rectangle with a height of 3. What is the volume of the object?"
This is the part of an exorbitant amount of problems that I have been working on that I don't know how to do and couldn't find any help for on Wikipedia or by searching google for teacher's notes and the like.

Anyways, even if you only want to tackle one problem, I would be greatly appreciative.

I also wanted to say that I have been shadowing this forum for a while, and would like to start contributing.

2. Originally Posted by Jon.Monreal
[snip]
1. "For what values of k does the graph of y=(4/3)x^3+2kx^2+5x+3 have two tangent lines parallel to the x-axis."
[snip]
You want dy/dx = 0 to have two solutions. Hint: A quadratic equation has two solutions when $\Delta > 0$.

Originally Posted by Jon.Monreal
[snip]
10. "Use differentials to estimate 4.1^4."
[snip]
You should know that $f(x+h) \approx f(x) + h f'(x)$. Consider the function $f(x) = x^{1/4}$, x = 4 and h = 1/10.

Originally Posted by Jon.Monreal
[snip]
"Find the volume of the solid formed when the region between y=tanx and the x-axis on the interval [0,(pi/4)] is revolved around the x-axis."
[snip]
Where exactly are you stuck here? Substitute into the usual volume of solid of revolution formula:

Rotation around x-axis: $V = \pi \int_a^b y^2 \, dx = \pi \int _{0}^{\pi/4} \tan^2 x \, dx$.

Most of the other questions are also basic application of formulae that should be in your class notes or textbook. It would help if you said where exactly you were stuck with each one.

3. Originally Posted by mr fantastic
You want dy/dx = 0 to have two solutions. Hint: A quadratic equation has two solutions when $\Delta > 0$.

You should know that $f(x+h) \approx f(x) + h f'(x)$. Consider the function $f(x) = x^{1/4}$, x = 4 and h = 1/10.

Where exactly are you stuck here? Substitute into the usual volume of solid of revolution formula:

Rotation around x-axis: $V = \pi \int_a^b y^2 \, dx = \pi \int _{0}^{\pi/4} \tan^2 x \, dx$.

Most of the other questions are also basic application of formulae that should be in your class notes or textbook. It would help if you said where exactly you were stuck with each one.
My problem is that I don't have a textbook, only a list of questions, and no resources at hand related to the problems. I'm pretty lost on these questions as to even setting up formulas.

EDIT: I have solved #9.

4. the formula for finding volume using washers is

If you still uncertain what to do ill work an example for you

5. Originally Posted by 11rdc11
the formula for finding volume using washers is

If you still uncertain what to do ill work an example for you
I'm just confused as to what the ' is for, and why it is on different sides of the perentheses.

6. I use the ' to indicate to the power of

for example 2'2 is (2 to the power of 2 which equals 4)

7. Originally Posted by 11rdc11
I use the ' to indicate to the power of

for example 2'2 is (2 to the power of 2 which equals 4)
Sounds good. But shouldn't this be a definite integral?

Perhaps an example would help. In the meanwhile, I'm going to continue trying to work at these.

8. Example: Find the volume of the solid of revolution obtained by revolving the plane region R bounded by y = sqrtx, the x axis, and the line x=4 about the x-axis

so the formula is

V= pi intergal (sqrtx)'2

V= pi intergal x

V=pi (1/2)x'2

so now plug in your upper bound of 4 and lower bound of 0

V= pi(16/2)

9. Yes it is a definite integral

10. Originally Posted by 11rdc11
Example: Find the volume of the solid of revolution obtained by revolving the plane region R bounded by y = sqrtx, the x axis, and the line x=4 about the x-axis

so the formula is

V= pi intergal (sqrtx)'2

V= pi intergal x

V=pi (1/2)x'2

so now plug in your upper bound of 4 and lower bound of 0

V= pi(16/2)
So, just to make sure I'm doing this right, for #6 I would end up with:

pi integral tan^2 x
pi(tanx-x)
pi(1-(pi/4)) which is about equal to 0.6741915533

Right?

11. Yep

12. Originally Posted by mr fantastic
You should know that $f(x+h) \approx f(x) + h f'(x)$. Consider the function $f(x) = x^{1/4}$, x = 4 and h = 1/10.
My question with this is, how did you get x=4 and h=1/10? Or are you referring to my problem and not the example?

I'm a bit confused.

EDIT: Solved #6 then. Thanks for checking it. By the way, when all of this is done, I am going to edit my first post so those who come back in the future can reference this thread.

13. 10. "Use differentials to estimate 4.1^4."

x=4 and dx=.1

y = x'4

dy= 4x'3dx

dy= 25.6

thus

x'4 +dy =

256 + 25.6 = 281.6

hope that helps

14. Originally Posted by 11rdc11
10. "Use differentials to estimate 4.1^4."

x=4 and dx=.1

y = x'4

dy= 4x'3dx

dy= 25.6

thus

x'4 +dy =

256 + 25.6 = 281.6

hope that helps
Okay, I understand now.

I was using my calculator to calculate the actual value (282.5751), and my answer wasn't even coming close. But that's just because I was doing the problem completely wrong.

That's it for #10 then. Right now I'm trying to work on #1. Anybody have anything?

15. Originally Posted by Jon.Monreal
On a good number of questions. I realize I'm new here, but I hope to stick around. I have left a more sizable message after the problems, in case you're interested.

Here goes:

:1. "For what values of k does the graph of y=(4/3)x^3+2kx^2+5x+3 have two tangent lines parallel to the x-axis."
This is the part of an exorbitant amount of problems that I have been working on that I don't know how to do and couldn't find any help for on Wikipedia or by searching google for teacher's notes and the like.

Anyways, even if you only want to tackle one problem, I would be greatly appreciative.

I also wanted to say that I have been shadowing this forum for a while, and would like to start contributing.

As mr fantastic suggested, you want to find where $\frac{\,dy}{\,dx}=0$

Since $y=\tfrac{4}{3}x^3+2kx^2+5x+3$, $\frac{\,dy}{\,dx}=4x^2+4kx+5$

To find where $\frac{\,dy}{\,dx}=0$, apply the quadratic formula:

$x=\frac{-4x\pm\sqrt{16k^2-80}}{8}$

However, to get 2 solutions, we need $16k^2-80>0$

Thus, $k<-\sqrt{5}$ and $k>\sqrt{5}$

Thus, $k$ can be any value within these two intervals:

$(-\infty,-\sqrt{5})$ and $(\sqrt{5},\infty)$

Note the round brackets. This means $k=\pm\sqrt{5}$ is not included in our answer set.

Does this make sense?

--Chris

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