Exercise 2.11Hyperbolic Functions and EquationsCircular & Hyperbolic Functionssin θ and cos θ are known as the the circular functions because: ┌──────────────────────┐
│ Parametric equations │
│ of a circle │
└───────────┬──────────┘
┌──────────┴─────────┐
x = cos θ, y = sin θ ──────────► x² + y² = 1
└─────┬─────┘
┌─────────────┴─────────────┐
│ equation of a circle │
├───────────────────────────┤
│ could also be written as: │
└─────────────┬─────────────┘
┌──────┴──────┐
y = ±√1 - x²
Which explains why, when we want to find integrals like these...⌠ 1 OR ⌠ ⌡ √1 - x² │ √1 - x² dx ⌡ ...the substitution: x = sin θ is needed!
But, if we want to integrate: ⌠
│ √x² - 1 dx
⌡
Then we need to consider the function that came from; and its parametric equations: y = √x² - 1
└─────┬─────┘
┌─────────────┴─────────────┐
│ could also be written as: │
└─────────────┬─────────────┘
┌──────┴──────┐
x² - y² = 1
OR
x = ½(t + 1/t), y = ½(t - 1/t) ◄───────── (x+y)(x-y) = 1
└───────────────┬───────────────┘
┌───────────┴──────────┐
│ parametric equations │
│ of a hyperbola ├─────────────────────────┐
└──────────────────────┘ ┌──────────────┴──────────────┐
│ not the same as the ones we │
│ used during conic sections! │
└─────────────────────────────┘
These equations become especially useful when we replace the parameter 't' with: ℮u:
Which reminds us a lot of:
So, we name these parametric equations as the hyperbolic cosine and the hyperbolic sine functions:
So now we can say:x = cos θ, y = sin θ ──────────► x² + y² = 1 x = cosh θ, y = sinh θ ──────────► x² - y² = 1 Other Hyperbolic FunctionsSince we see that cosh u and sinh u are very similar in definition to cos θ and sin θ, it is easy for us to define:
The hyperbolic functions share a lot of similarities with the circular functions:
Solving Hyperbolic Equationse.g. Solve: cosh u = 2 sinh u - 1 We have a couple of options here:
1) Refer back to the original definitions of cosh u and sinh u:
cosh u = 2 sinh u - 1
└──┬───┘ └───┬──┘
┌────┴───┐ ┌────┴───┐
½(℮u+℮-u) = 2{½(℮u-℮-u)} - 1
Replace ℮u with x: ½(x + 1/x) = 2{½(x - 1/x)} - 1
x + 1/x = 2{(x - 1/x)} - 2
×x ×x ×x ×x ×x
------------------------------
x² + 1 = 2x² - 2 - 2x
0 = x² - 2x - 3
0 = (x - 3)(x + 1)
x = 3, x = -1
└──┬──┘ └──┬───┘
┌────┘ └────┐
┌───┴──┐ ┌───┴───┐
Replace back, x with ℮u: ℮u = 3 ℮u = -1
ln ln ln ln
-------- -------
u = ln 3 N/A
2) Use the Compound Angle Transformation:
Re-arrange the equation to: 1 = 2 sinh u - cosh u
└─────────┬─────────┘
│ │ To find artanh(½), solve: │
└─────┬────────────┤ ℮2u - 1 = 1 │
│ │ ℮2u + 1 = 2 │
┌─┴─┐
1 = R sinh(u - ½ln3)
▲ ┌────────────────────────────────────┐
└─────────────────────┤ Of course, it is not: R² = A² + B² │
│ it is: R² = A² - B² │
└────────────────────────────────────┘
1 = √3 sinh(u - ½ln3)
1/√3 = sinh(u - ½ln3)
sinh-1 sinh-1 ◄────────────────────- ┤ sinh u = 1/√3 │
------------------------- │ └───┬──┘ │
│ ┌─────┴──┐ │
┌───────────────────────────────────-┤ ½(℮u-℮-u) = 1/√3 │
┌──┴──┐
½ln3 = u - ½ln3
ln 3 = u
e.g. Solve: 12 cosh²u + 7 sinh u = 2412 cosh²u + 7 sinh u = 24 └──┬───┘ ┌─────────┴──────────┐ │ Replace the cosh²u ├────────────────┤ since the ID is: cosh²u - sinh²u = 1 │ │ with 1 + sinh²u │ └─────────┬──────────┘ ┌────┴────┐ 12(1 + sinh²u) + 7 sinh u = 24 12 sinh²u + 7 sinh u - 24 = 0 (3 sinh u + 4)(4 sinh u - 3) = 0 sinh u = -4/3 or sinh u = 3/4 └───┬──┘ └───┬──┘ ┌──────┴─┐ ┌─────┴──┐ ½(℮u-℮-u) = -4/3 ½(℮u-℮-u) = 3/4 I've omitted the working: u = ln 1/3 u = ln 2 Question 1, part (a): There are two ways we can go about this: If we replace cosh(2x) with: ½(℮2x + ℮-2x), and then use Maclaurin's Method: f(x) = ½(℮2x + ℮-2x) ──────────────────── ┼──► f(0) = 1 ───────────►─────────────────────┐
│ │
---------------------------------------------- ┼------------------------------------------------ │
f’(x) = ½(2℮2x - 2℮-2x) ─────────────────── ┼──► f’(0) = 0 ───────────►─────────────────────┼─┐
│ │ │
---------------------------------------------- ┼------------------------------------------------ │ │
f’’(x) = ½(4℮2x + 4℮-2x) ─────────────────── ┼──► f’’(0) = 4 ───────────►─────────────────────┼─┼─┐
│ │ │ │
---------------------------------------------- ┼------------------------------------------------ │ │ │
f’’’(x) = ½(8℮2x - 8℮-2x) ─────────────────── ┼──► f’’’(0) = 0 ───────────►─────────────────────┼─┼─┼─┐
│ │ │ │
┌─────────────────────────────────────◄──────────────────────────────────┘ │ │ │
│ ┌──────────────────────────────◄─────────────────────────────────┘ │ │
│ │ ┌─────────────────────◄────────────────────────────────┘ │
│ │ │ ┌──────────◄───────────────────────────────┘
┌─┴─┐ ┌──┴─┐ ┌──┴──┐ ┌──┴───┐
f(x) ≈ f(0) + f’(0) x + f’’(0) x2 + f’’’(0) x3 + f(4)(0) x4 + ...
1! 2! 3! 4! .
Alternatively, we could have just substituted '2x' into the expasion of cosh u: cosh u = 1 + 1u2 + 1u4 + ... 2! 4! cosh(2x) = 1 + 1(2x)2 + 1(2x)4 + ... 2! 4! Question 1, part (b): If we replace sinh(½x) with: ½(℮½x - ℮-½x), and then use Maclaurin's Method: f(x) = ½(℮½x - ℮-½x) ──────────────────── ┼──► f(0) = 0 ───────────►─────────────────────┐
│ │
---------------------------------------------- ┼------------------------------------------------ │
f’(x) = ½(½℮½x + ½℮-½x) ─────────────────── ┼──► f’(0) = ½ ───────────►─────────────────────┼─┐
│ │ │
---------------------------------------------- ┼------------------------------------------------ │ │
f’’(x) = ............ ───────────────────── ┼──► f’’(0) = 0 ───────────►─────────────────────┼─┼─┐
│ │ │ │
---------------------------------------------- ┼------------------------------------------------ │ │ │
f’’’(x) = ............ ───────────────────── ┼──► f’’’(0) = … ───────────►─────────────────────┼─┼─┼─┐
│ │ │ │
┌─────────────────────────────────────◄──────────────────────────────────┘ │ │ │
│ ┌──────────────────────────────◄─────────────────────────────────┘ │ │
│ │ ┌─────────────────────◄────────────────────────────────┘ │
│ │ │ ┌──────────◄───────────────────────────────┘
┌─┴─┐ ┌──┴─┐ ┌──┴──┐ ┌──┴───┐
f(x) ≈ f(0) + f’(0) x + f’’(0) x2 + f’’’(0) x3 + f(4)(0) x4 + ...
1! 2! 3! 4! .
Alterntively, we could have just... Question 1, part (c): Instead of using Maclaurin, we could just take our existing expansions (for cosh x and for sinh x) and add them together: cosh x = 1 + 1x2 + 1x4 + ... 2! 4! sinh x = 1x + + 1x3 + 1x5 + ... 1! 3! 5! Alternatively, we could have...
Question 1, part (d): Starting with: f(x) = ⎧ ℮x + ℮-x ⎫2
⎪ 2 ⎪
⎩ ⎭
It is probably easier to do all the differentation we are going to need if we simplify this first: f(x) = { ℮x + ℮-x }2
4 .
f(x) = ¼(℮x + ℮-x)(℮x + ℮-x )
f(x) = ¼(℮2x + 2 + ℮-2x) ──────────► f(0) = 1
So, differentiating: f’(x) = ¼(2℮2x - 2℮-2x) ──────────► f’(0) = 0 Etc...
Question 1, part (f): I thunk iz ezyier if you simplifii dis firs
Question 1, part (g): This won't really simplify - so we just need to get on with differentiating: f(x) = ln(cosh x) ─────────────── ┼──► f(0) = 0 ───────────►─────┐
│ │
-------------------------------------- ┼-------------------------------- │
f’(x) = tanh x ─────────────────── ┼──► f’(0) = 0 ───────────►─────┼─┐
│ │ │
-------------------------------------- ┼-------------------------------- │ │
f’’(x) = sech²x ─────────────────── ┼──► f’’(0) = 1 ───────────►─────┼─┼─┐
│ │ │ │
-------------------------------------- ┼-------------------------------- │ │ │
f’’’(x) = -2f’’(x)f’(x) ──────────── ┼──► f’’’(0) = … ───────────►─────┼─┼─┼─┐
│ │ │ │
┌─────────────────────────────────────◄──────────┘ │ │ │
│ ┌──────────────────────────────◄─────────┘ │ │
│ │ ┌─────────────────────◄────────┘ │
│ │ │ ┌──────────◄───────┘
┌─┴─┐ ┌──┴─┐ ┌──┴──┐ ┌──┴───┐
f(x) ≈ f(0) + f’(0) x + f’’(0) x2 + f’’’(0) x3 + f(4)(0) x4 + ...
1! 2! 3! 4! .
Note: When we differentiate ${ sech }^{ 2 }x$, we get: $-2{ sech }^{ 2 }x \tanh { x }$, which can be re-written as: $-2f\prime\prime \left( x \right) f\prime \left( x \right)$ From then on, (becuase we've managed to write $f\prime \prime\prime \left( x \right)$ in terms of $f\prime \prime \left( x \right)$ and $f\prime \left( x \right)$) the rest of the differentiation becomes as simple as popping a pimple...
Question 1, part (h): Again, this won't simplify, so we'll just have to differentiate - a lot!
Question 2: So they want us to re-write $\cosh ^{ 2 }{ x }$ (which is a POWER of $\cosh { x }$) in terms of MULTIPLE ANGLES... Remember - we used to do that sort of thing when we were integrating even powers of $\sin { x }$ or $\cos { x }$... And the WAY we did it, was using the DOUBLE-ANGLE IDs
So, we can re-write: $\left\{ \color{maroon}{\cosh ^{ 2 }{ x }} \right\} ^{ 2 } \; =\; {\left\{ \color{maroon}{\frac { 1 }{ 2 } \left( 1+\cosh { 2x } \right)} \right\} }^{ 2 }$
Question 3, part (a): Since we already know he expansion of sinh u, we can find the expansion of sin x² by simply substituting every ‘u’ in the expansion, with ‘x²’
Question 3, part (b): It says, 'USE ANSWER TO (A)', so you are just integrating: $\int _{ 0 }^{ 0.6 }{ { x }^{ 2 } } \left( { x }^{ 2 }+\frac { 1 }{ 6 } { x }^{ 6 }+\frac { 1 }{ 120 } { x }^{ 10 } \right) \; dx$ Expanding, integrating and shoving in the limits...
Question 4: They want us to go back to the basic definitions: $\cosh { x } \;=\;\frac { 1 }{ 2 } \left( { e }^{ x }\;+\;{ e }^{ -x } \right)$ and $\sinh { x } \; =\; \frac { 1 }{ 2 } \left( { e }^{ x }\; -\; { e }^{ -x } \right)$ in order to prove these IDs: Part (a): Thinking about the expnential forms - effectively, they want us to prove: $${ \left( \frac { { e }^{ x }+{ e }^{ -x } }{ { e }^{ x }-{ e }^{ -x } } \right) }^{ 2 }\; -\; 1\quad \equiv \quad { \left( \frac { 2 }{ { e }^{ x }-{ e }^{ -x } } \right) }^{ 2 }$$ Which isn't particularly hard (Clue: Do common denominator for the L.H.S, but don't expand the bottom...) Part (b): Again, in exponential terms, this means we gotta prove: $$\frac { 1 }{ 2 } \left( { e }^{ x+y } \;-\; { e }^{ -\left( x+y \right) } \right) \quad \equiv \quad \frac { 1 }{ 2 } \left( { e }^{ x }\; -\; { e }^{ -x } \right) \cdot \frac { 1 }{ 2 } \left( { e }^{ y }\; +\; { e }^{ -y } \right) \;+\;\frac { 1 }{ 2 } \left( { e }^{ x }\; +\; { e }^{ -x } \right) \cdot \frac { 1 }{ 2 } \left( { e }^{ y }\; -\; { e }^{ -y } \right)$$ Again, this ain't hard - just expand the R.H.S (but leave the $\frac { 1 }{ 4 }$ outside)... Part (c): Exponentially, we are trying to prove: $$\frac { { e }^{ 2x }-{ e }^{ -2x } }{ { e }^{ 2x }+{ e }^{ -2x } } \quad \equiv \quad \frac { 2\left( \frac { { e }^{ x }-{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } } \right) }{ 1 \;+\; { \left( \frac { { e }^{ x }-{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } } \right) }^{ 2 } }$$ Which I can forsee getting rather messy, unless you work carefully on simplifying the R.H.S. Part (d): D'ya remember this - Ex 2.5a, qu 16; and I told you, if you remember the expansions of $\sin { 3\theta }$ and $\cos { 3\theta }$, then one day, it might come in handy. You swore at me right? We'll guess who's laughing now? So, we need to prove: $$\frac { 1 }{ 2 } \left( { e }^{ 3x }+{ e }^{ -3x } \right) \quad \equiv \quad 4{ \left\{ \frac { 1 }{ 2 } \left( { e }^{ x }+{ e }^{ -x } \right) \right\} }^{ 3 }-\;3\left\{ \frac { 1 }{ 2 } \left( { e }^{ x }+{ e }^{ -x } \right) \right\}$$ Clue: It might help you to remember that ${ a }^{ 3 }+{ b }^{ 3 }=\left( a+b \right) \left( { a }^{ 2 }-ab+{ b }^{ 2 } \right)$
Question 5: We've got to ‘guess’ the equivalent hyperbolic identity - and then we've got to prove it! Part (a): There is no $\sin ^{ 2 }{ \theta }$ in this Identity, so the hyperbolic version will be the same: $$\sinh { \left( x+y \right)} \quad \equiv \quad \sinh { x } \cosh { y } \;+\;\cosh { x } \sinh { y }$$ And, we already proved this in Question 4, part (b), so 😏 Part (b): This time the $\sin { x } \sin { y }$ is the equivalent of a $\sin ^{ 2 }{ \theta }$, the the $SIGN$ in-front of this term will change in the hyperbolic version: $$\cosh { \left( x-y \right) } \quad \equiv \quad \cosh { x } \cosh { y } \; -\; \sinh { x } \sinh { y }$$ The proof will owrk in almost the same way as in question 4, part (b)... Part (c): The last term in this ID ($-4\sin ^{ 3 }{ x }$) effectively contains a $\sin ^{ 2 }{ x }$ term, so the $SIGN$ in-front of this term will change in the hyperbolic version... And we did a similar proof in question 4, part (d)... Part (d): There is no $\sin ^{ 2 }{ x }$ term in this FACTOR FORMULA, so the hpyerbolic version will be the same... $$\cosh { x } \;+\;\cosh { y } \quad \equiv \quad 2\cosh { \left( \frac { x+y }{ 2 } \right) } \cosh { \left( \frac { x-y }{ 2 } \right) }$$ So, we need to prove: $$\frac { 1 }{ 2 } \left( { e }^{ x }+{ e }^{ -x } \right) \; +\; \frac { 1 }{ 2 } \left( { e }^{ y }+{ e }^{ -y } \right) \quad \equiv \quad 2\cdot \frac { 1 }{ 2 } \left( { e }^{ \frac { x+y }{ 2 } }+{ e }^{ \frac { -\left( x+y \right) }{ 2 } } \right) \cdot \frac { 1 }{ 2 } \left( { e }^{ \frac { x-y }{ 2 } }+{ e }^{ \frac { y-x }{ 2 } } \right)$$ Multiply out the R.H.S. and it tidies up nicely...
Question 6, part (a): There are two ways of doing this: We could solve sinh x = 3/4 to find x sinh x = ¾ └───┬──┘ ┌──────┴─┐ ½(℮x-℮-x) = ¾ Replacing ℮x with u: ½(u - 1/u) = ¾ ──────────► 2u² - 3u - 2 = 0 (2u+…)(u-…) = 0 └──┬─┘└──┬───┘ ┌────┘ └────┐ ┌───┴──┐ ┌───┴──┐ Replace back, u with ℮x: ℮x = -… ℮x = 2… ln ln ln ln -------- -------- N/A x = ln… And then put this into the eqution for cosh u to find cosh x (and recip to find sech x), divide sinh x by cosh x to find tanh x etc... Alternatively, we can use the Pythag IDs: cosh²x - sinh²x = 1 └──┬───┘ ┌─┴─┐ cosh²x - ¾ = 1 –› cosh x = … Then recip this answer to find: sech x
Question 6, part (b): Again, it is probably easier to use the Hyperbolic ID connecting $\tanh ^{ 2 }{ x }$ and ${ sech }^{ 2 }x$, work out $sech\;x$ then you'll know $\cosh { x }$ and life will be sweet...
Question 6, part (c): Okay - there's a difference between this part, and parts (a) & (b). You know what that is init?
Question 8: We could do this in one of two ways: 1) Using the Hyperbolic IDs: $$\begin{align} \tanh{x} \;= & \sqrt { \frac { \color{green}{\cosh { 2x }} -1 }{ \color{teal}{\cosh { 2x }} +1 } } \\ = & \sqrt { \frac { \color{green}{1+2\sinh ^{ 2 }{ x }} -1 }{ \color{teal}{...........}+1 } } \\ = & \sqrt { \frac { 2\sinh ^{ 2 }{ x } }{ ........... } } \end{align}$$ 2) By going back to the exponential definitions: $$\begin{align} \frac { { e }^{ x }-{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } } \; = & \sqrt { \frac { \color{green}{\frac { 1 }{ 2 } \left( { e }^{ x }+{ e }^{ -x } \right)} -1 }{ \color{teal}{\frac { 1 }{ 2 } \left( { e }^{ x }+{ e }^{ -x } \right)} +1 } } \\ = & \sqrt { \frac { { e }^{ x }+{ e }^{ -x }-2 }{ ...........+2 } } \\ = & \sqrt { \frac { { \left( { e }^{ x }-{ e }^{ -x } \right) }^{ 2 } }{ ........... } } \end{align}$$
Question 9: Again, we have two choices: 1) Using hyperbolic IDs: You need to replace $\cosh { 3x }$ with $4\cosh ^{ 3 }{ x } -3\cosh { x }$, and replace $\sinh { 3x }$ with $3\sinh { x } + 4\sinh ^{ 3 }{ x }$ and then do your magic... 2) By going back to the exponential definitions: So, we need to prove: $$\frac { 1 }{ 8 } { \left( { e }^{ 2x }+{ e }^{ -2x } \right) }^{ 3 }\quad \equiv \quad \frac { 1 }{ 2 } \left( { e }^{ 3x }+{ e }^{ -3x } \right) \cdot \frac { 1 }{ 8 } { \left( { e }^{ x }+{ e }^{ -x } \right) }^{ 3 }\; +\; \frac { 1 }{ 2 } \left( { e }^{ 3x }-{ e }^{ -3x } \right) \cdot \frac { 1 }{ 8 } { \left( { e }^{ x }-{ e }^{ -x } \right) }^{ 3 }$$
Question 10: Easy peasy - just substitute: $x\;=\;a\cosh { x }$ and $y\;=\;......$ into your left and right eyes...
Question 11, part (a): We can replace the Hyperbolic functions with their definitions: sinh x + 4 = 4 cosh x
└──┬───┘ └───┬──┘
┌────┴───┐ ┌────┴───┐
½(℮x-℮-x) + 4 = 4{½(℮u+℮-u)}
Replace ℮u with x: ½(x - 1/x) + 4 = 2(x - 1/x)
x + 1/x + 8 = 4(x - 1/x)
×x ×x ×x ×x ×x
------------------------------
x² + 1 + 8x = 4x² - 4
0 = 3x² - 8x - 3
x = …, x = …
└──┬──┘ └──┬──┘
┌────┘ └────┐
┌───┴──┐ ┌───┴───┐
Replace back, x with ℮u: ℮u = … ℮u = …
ln ln ln ln
-------- -------
u = ln … N/A
Question 11, part (e): This time - it's the double angle ID we need to use, so replace the $\cosh { \left( 2\x \right) }$ with either $2\cosh ^{ 2 }{ \theta } \; -\; 1$ or with $1\; +\; 2\sinh ^{ 2 }{ \theta }$
Question 11, part (f): Easy - replace the $\tanh ^{ 2 }{ x }$ with $1-{ sech }^{ 2 }x$
Question 11, part (g): You know that $\cosh { x }$ has an exponential derivation? Unlike, you - who has an unknown derivation...
Question 12: Okay, you know a 'QUICK-WAY' of doing the Compound Angle Transformation for the Circular Functions, but DON'T ASSUME the method will be exactly the same for Hyperbolic Functions... Use the slow method and get it right: Step 1: Expand the compound angle:
sinh (x + α) = sinh x cosh α + cosh x sinh α
R sinh (x + α) = R sinh x cosh α + R cosh x sinh α
Step 2: Underneath this, line up the terms of the expression you want to factorise so that the
'sin θ' term is under the 'sin θ' and the 'cos θ' term is under the 'cos θ':
R sinh (x + α) = R sinh x cosh α + R cosh x sinh α
5 sinh x + 4 cosh x
Step 3: If you cross out the 'matching' bit of each term:
R sinh (x + α) = R
So, you are wondering - is there a quick way for CATH (Hyberbolic CAT)? Yes - here are the rules for applying the CATH to $y\;=\;A\cosh { x } +B\sinh { x }$: 1) If $A\;>\;B$, then use: $R\cosh { \left( x+\alpha \right) }$,
but if $B\;>\;A$, then use: $R\sinh { \left( x+\alpha \right) }$ 2) To find $R$ and $\alpha$, use: ${ R }^{ 2 }\;=\;\left| { A }^{ 2 }\;-\;{ B }^{ 2 } \right|$ and $\alpha \;=\;\tanh ^{ -1 }{ \frac { B }{ A } }$ $\alpha \;=\;\tanh ^{ -1 }{ \frac { A }{ B } }$ 3) You can find $\tanh ^{ -1 }{ t }$ using $\tanh ^{ -1 }{ t } \;=\;\frac { 1 }{ 2 } \ln { \left( \frac { 1+t }{ 1-t } \right) }$ or by just sticking it in your calculator...
Question 14: The graph $y \;=\; \cosh { x }$ has a minimum value, but $y \;=\; \sinh { x }$ doesn't. So apply CATH and then use transformations to sketch your head!
Question 15, part (a): Replace the $\coth { x }$ with $\frac{ 1 }{\tanh{ x }}$ and replace your bed with a cot!
Question 15, part (b): Okay - nothing is square (apart from you), so I'm going back to exponentials...
Question 15, part (c): Much better: Replace the ${sech}^{2}x$ with $1 - \tanh^{2}{x}$
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