Solving a Complex Math Problem: Help Appreciated

[Nemo1]

Hi Community,

I have the following problem and I would like some help in understanding part a.
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So far I far I have been able to show that:

$$1+\frac{(e^x-e^{-x})^2}{4}$$ = $$\frac{(e^x)^2-2(e^x-e^{-x})+(e^{-x})2}{4}+1$$

But I am unsure of how to proceed.

Also any pointers on how to look at the other two parts would be appreciated.

Finding the length of the curve I am thinking I need to use an arc length formula, but unsure of which?

For finding the volume I am thinking I can use $$\pi\int_{a}^{b}r^2 \,dx$$ but need to understand how to find $f(x)$ from $x=0$ & $x=1$ from the formula as I am thinking that the $y$ value should be my radius.

Many thanks for your time in advance.

Cheers Nemo.

 

[MarkFL]

For part a), you are not expanding the given square correctly...recall:

$$(a+b)^2=a^2+2ab+b^2$$

Let's write:

$$\left(\frac{e^x-e^{-x}}{2}\right)^2=\frac{1}{4}\left(e^x-e^{-x}\right)^2$$

What do you get now when expanding the remaining square?

 

[Nemo1]

For part a), you are not expanding the given square correctly...recall:

$$(a+b)^2=a^2+2ab+b^2$$

Let's write:

$$\left(\frac{e^x-e^{-x}}{2}\right)^2=\frac{1}{4}\left(e^x-e^{-x}\right)^2$$

What do you get now when expanding the remaining square?

Would it be?

$$\frac{1}{4}(e^x)^2+2(e^x)(-e^{-x})+(-e^{-x})^2$$

 

[MarkFL]

Would it be?

$$\frac{1}{4}(e^x)^2+2(e^x)(-e^{-x})+(-e^{-x})^2$$

Yes...can you simplify further?

 

[Nemo1]

Yes...can you simplify further?

So I have been working on this and must admit I am overwhelmed!

I have found that we are missing a $1+$ to the front of our formula to be equal to the original $$1+({\frac{e^{x}-e^{-x}}{2}})^2$$

By using the exponent rule $$a^b\cdot a^c=a^{b+c}$$

I can simplify the centre to:

$$1+\frac{(e^{x})^2-2+(-e^{-x})^2}{4}$$

Rearranging:

$$1+\frac{(e^{x})^2+(-e^{-x})^2-2}{4}$$

From here I am totally stumped, Sorry Mark, I really am trying to understand this!

Cheers Nemo(Sweating)

 

[MarkFL]

Yes, I wanted you to focus on the expansion of the square first, and then we would add the $1$. You simplified correctly, and so what we have now is:

$$1+\frac{(e^{x})^2+(-e^{-x})^2-2}{4}$$

Get a common denominator and add those two terms, and simplify by combining like terms in the numerator...what do you have?

 

[Nemo1]

Yes, I wanted you to focus on the expansion of the square first, and then we would add the $1$. You simplified correctly, and so what we have now is:

$$1+\frac{(e^{x})^2+(-e^{-x})^2-2}{4}$$

Get a common denominator and add those two terms, and simplify by combining like terms in the numerator...what do you have?

Slight light bulb moment:

$$1+\frac{(e^{x})^2+(-e^{-x})^2-2}{4}$$

Can become:

$$\frac{4}{4}+\frac{(e^{x})^2+(-e^{-x})^2-2}{4}$$

Then can become:

$$\frac{4+(e^{x})^2+(-e^{-x})^2-2}{4}$$

Then:

$$\frac{(e^{x})^2+(-e^{-x})^2+2}{4}$$

Then:

$$\frac{(e^{x})(e^{x})+(-e^{-x})(-e^{-x})+2}{4}$$

Then:

$$\frac{(e^{x})(e^{x})-e^{-x}(-e^{-x})+2}{4}$$

Then:

$$\frac{e^{2x}--e^{-2x}+2}{4}$$

Then:

$$\frac{e^{2x}+e^{-2x}+2}{4}$$

What do you recommend from here?
Or should I have gone a different route on a prior step?

Cheers Nemo

 

[MarkFL]

You have correctly found:

$$\frac{e^{2x}+e^{-2x}+2}{4}$$

Let's arrange this as:

$$\frac{e^{2x}+2+e^{-2x}}{4}$$

Now, recalling that $$1=e^xe^{-x}$$

Can we rewrite that middle term so that we can then apply the formula for the square of a binomial? (Thinking)

 

[Nemo1]

You have correctly found:

$$\frac{e^{2x}+e^{-2x}+2}{4}$$

Let's arrange this as:

$$\frac{e^{2x}+2+e^{-2x}}{4}$$

Now, recalling that $$1=e^xe^{-x}$$

Can we rewrite that middle term so that we can then apply the formula for the square of a binomial? (Thinking)

$$\frac{e^{2x}+2+e^{-2x}}{4}$$

Can become:

$$\frac{e^{2x}+2(e^xe^{-x})+e^{-2x}}{4}$$

Which is in the format of:

$$(a+b)^2=A^2+2ab+b^2$$

Which then gives us:

$$\frac{(e^{x}+e^{-x})^2}{4}$$

Which is finally equal to:

$$(\frac{e^{x}+e^{-x}}{2})^2$$

which is the function we are trying to get.

Wow! I would not have ever realized that $$1=e^xe^{-x}$$

Thanks Mark, now onto attempting the next bit.

 

[Nemo1]

Hi Mark, to find the length,

I think its the following,

Using the formula:

$$Length=\int_{a}^{b}\sqrt{1+f'(x)^2} \,dx$$

Knowing that the derivative of $$\frac{e^{x}+e^{-x}}{2}$$ = $$\frac{1}{2}(e^{-x}+e^{x})$$

We can plug this into get:

$$Length=\int_{a}^{b}\sqrt{1+(\frac{1}{2}(e^{-x}+e^{x}))^2} \,dx$$

Which without showing the working out is equal to $$\frac{e^2-1}{2e}=\approx 1.1752$$

Do you concur?

Cheers Nemo

 

[MarkFL]

Okay for part b), I would write:

$$f(x)=\frac{e^x+e^{-x}}{2}$$

$$f'(x)=\frac{e^x-e^{-x}}{2}$$

So the arc-length $s$ is given by

$$s=\int_0^1 \sqrt{1+[f'(x)]^2}\,dx$$

We found in part a) that:

$$1+[f'(x)]^2=\left(\frac{e^x+e^{-x}}{2}\right)^2$$

And so we have:

$$s=\frac{1}{2}\int_0^1 e^x+e^{-x}\,dx=\frac{1}{2}\left[e^x-e^{-x}\right]_0^1=\frac{1}{2}\left(\left(e-\frac{1}{e}\right)-(1-1)\right)=\frac{e^2-1}{2e}$$

So yes, I concur. :D

 

[Nemo1]

For Part C:

Using the formula for volume around an axis:

$$\pi\int_{a}^{b}r^2 \,dx$$

I can set up my integral as:

$$\pi\int_{0}^{1}(\frac{e^{x}+e^{-x}}{2})^2 \,dx$$

Simplifying and taking the indefinite integral (Take too long to type in my working out):

$$\pi \int \frac{1}{4}(e^{x}+e^{-x})^2 \,dx$$ = $$\frac{1}{2}\pi(x+sinh(2x))+c$$

Then I can calculate the boundaries using the (F.T.O.C)

$$\frac{1}{2}\pi(1+sinh(2\cdot1))-\frac{1}{2}\pi(0+sinh(2\cdot0))$$

Which is equal to:

$$\frac{1}{4}\pi(2+sinh(2)) = \approx 4.41932583$$

I think I am getting a handle on calculating integrals, but I must admit, that I find it difficult to know when to use substitution or when there is a suitable trig id available and or how to manipulate the functions so I can use one.

Practice, practice, practice I suppose.

Many thanks for your teaching Mark!

Cheers Nemo

 

[MarkFL]

I will check your result with the shell method. We have:

$$y=\frac{e^x+e^{-x}}{2}$$

$$2y=e^x+e^{-x}$$

$$2ye^x=e^{2x}+1$$

Arranging as a quadratic in $e^x$, we get:

$$\left(e^{x}\right)^2-2ye^x+1=0$$

And so, by the quadratic formula, we get:

$$e^x=\frac{-(-2y)\pm\sqrt{(-2y)^2-4(1)(1)}}{2(1)}=y\pm\sqrt{y^2-1}$$

Now, observing that:

$$y(1)=\frac{e^2+1}{2e}$$

we find:

$$e^1=\frac{e^2+1}{2e}\pm\sqrt{\left(\frac{e^2+1}{2e}\right)^2-1}$$

$$2e^2=\left(e^2+1\right)\pm\sqrt{\left(e^2+1\right)^2-4e^2}$$

$$2e^2=\left(e^2+1\right)\pm\left(e^2-1\right)$$

From this, we can see we want the larger of the two quadratic roots:

$$e^x=y+\sqrt{y^2-1}$$

Hence:

$$x=\ln\left(y+\sqrt{y^2-1}\right)$$

And so we may state:

$$V=\pi+2\pi\int_{y(0)}^{y(1)} u\left(1-\ln\left(u+\sqrt{u^2-1}\right)\right)\,du=\pi\left(1+\frac{1}{4}(\sinh(2)-2)\right)=\frac{\pi}{4}\left(\sinh(2)+2\right)\checkmark$$

It appears the two methods agree. (Yes)