CP-8 Hyperbolic functions — printable coverage pack

CP-8.1 · Understand the definitions of hyperbolic functions sinh x, cosh x and tanh x, including their domains and ranges, and be able to sketch their graphs.

$\sinh x=\dfrac{e^x-e^{-x}}2$ , $\cosh x=\dfrac{e^x+e^{-x}}2$ and $\tanh x=\dfrac{\sinh x}{\cosh x}$ .
All three functions have domain $\mathbb R$ ; their ranges are $\mathbb R$ , $[1,\infty)$ and $(-1,1)$ respectively.
$\sinh x$ and $\tanh x$ are odd and increasing, while $\cosh x$ is even with minimum point $(0,1)$ ; $y=\pm1$ are asymptotes of $y=\tanh x$ .
Do not give $\tanh x$ the closed range $[-1,1]$ : its graph approaches but never reaches either horizontal asymptote.

Tier 1 · Easy

1. State the domain, range and $y$ -intercept of $y=\cosh x$ .[3 marks]
Answer
- Domain $\mathbb R$ ; range $[1,\infty)$ ; $y$ -intercept $(0,1)$ .
Method: The exponential definition is valid for every real $x$ . Also $\cosh x\geq1$ , with equality at $x=0$ , so the graph crosses the $y$ -axis at $(0,1)$ .

Tier 2 · Standard

1. Using the exponential definitions, find the exact values of $\sinh(\ln3)$ , $\cosh(\ln3)$ and $\tanh(\ln3)$ .[4 marks]
Answer
- $\sinh(\ln3)=\dfrac43$ , $\cosh(\ln3)=\dfrac53$ , $\tanh(\ln3)=\dfrac45$ .
Method: Since $e^{\ln3}=3$ and $e^{-\ln3}=1/3$ , $\sinh(\ln3)=\frac12(3-1/3)=4/3$ and $\cosh(\ln3)=\frac12(3+1/3)=5/3$ . Their quotient is $\tanh(\ln3)=4/5$ .

Tier 3 · Hard

1. Sketch $y=2-3\tanh x$ . Label its intercept and both horizontal asymptotes, and state its range.[5 marks]
Answer
- A decreasing sigmoid through $(0,2)$ with asymptotes $y=5$ as $x\to-\infty$ and $y=-1$ as $x\to\infty$ .
- Range $(-1,5)$ .
Method: The graph of $\tanh x$ increases from $-1$ to $1$ and passes through the origin. Multiplication by $-3$ reverses it and scales vertically; adding $2$ gives the intercept $(0,2)$ . Transforming the limiting values gives $2-3(-1)=5$ and $2-3(1)=-1$ , neither attained.

CP-8.2 · Differentiate and integrate hyperbolic functions.

The basic derivatives are $(\sinh x)'=\cosh x$ , $(\cosh x)'=\sinh x$ and $(\tanh x)'=\operatorname{sech}^2x$ .
Apply the chain, product and quotient rules exactly as for other differentiable functions; the derivative of $\cosh x$ has no minus sign.
Reverse these results when integrating, including the inner derivative: for example $\int\sinh(ax+b)\,dx=\dfrac1a\cosh(ax+b)+C$ .
A common error is to copy circular-trigonometric signs, writing $(\cosh x)'=-\sinh x$ instead of $+\sinh x$ .

Tier 1 · Easy

1. Differentiate $y=5\cosh(4x)$ with respect to $x$ .[2 marks]
Answer
- $\dfrac{dy}{dx}=20\sinh(4x)$
Method: Differentiate $\cosh(4x)$ to obtain $4\sinh(4x)$ by the chain rule, then multiply by $5$ .

Tier 2 · Standard

1. Find $\displaystyle\int(5\cosh(3x)-2\sinh x)\,dx$ .[3 marks]
Answer
- $\dfrac53\sinh(3x)-2\cosh x+C$
Method: Reverse the chain rule: $\int5\cosh(3x)\,dx=(5/3)\sinh(3x)$ . Also $\int-2\sinh x\,dx=-2\cosh x$ . Add the arbitrary constant.

Tier 3 · Hard

1. The function is $f(x)=e^{-x}\cosh(2x)$ . Find the exact coordinate of its stationary point and determine its nature.[6 marks]
Answer
- $x=\dfrac14\ln3$ , $f(x)=\dfrac{2}{3^{3/4}}$ ; the stationary point is a minimum.
Method: Product differentiation gives $f'(x)=e^{-x}[2\sinh(2x)-\cosh(2x)]$ . Hence $f'(x)=0$ when $\tanh(2x)=1/2$ , so $2x=\frac12\ln3$ and $x=\frac14\ln3$ . At this value, $e^{2x}=3^{1/2}$ , so $\cosh(2x)=2/\sqrt3$ and $e^{-x}=3^{-1/4}$ ; thus $f=2/3^{3/4}$ . Since $2\tanh(2x)-1$ changes from negative to positive, the point is a minimum.

CP-8.3 · Understand and be able to use the definitions of the inverse hyperbolic functions and their domains and ranges.

$y=\operatorname{arsinh}x$ means $x=\sinh y$ ; it has domain $\mathbb R$ and range $\mathbb R$ .
$y=\operatorname{arcosh}x$ is the inverse of the restriction of $\cosh$ to $x\geq0$ , so its domain is $[1,\infty)$ and its range is $[0,\infty)$ .
$y=\operatorname{artanh}x$ means $x=\tanh y$ ; it has domain $(-1,1)$ and range $\mathbb R$ .
Do not treat $\cosh$ as one-to-one on all real numbers: its principal inverse returns the non-negative value only.

Tier 1 · Easy

1. State the domain and range of $y=\operatorname{arcosh}x$ .[2 marks]
Answer
- Domain $[1,\infty)$ ; range $[0,\infty)$ .
Method: The inverse uses the one-to-one branch of $\cosh y$ for $y\geq0$ . That branch takes values from $1$ upwards.

Tier 2 · Standard

1. Solve $\operatorname{arsinh}(2x-1)=\ln2$ exactly.[3 marks]
Answer
- $x=\dfrac78$
Method: Apply $\sinh$ to both sides: $2x-1=\sinh(\ln2)=\frac12(2-1/2)=3/4$ . Hence $2x=7/4$ and $x=7/8$ .

Tier 3 · Hard

1. Let $u=\operatorname{artanh}(1/3)$ . Find the exact value of $x$ satisfying $\operatorname{arcosh}x=2u$ , without using decimal approximations.[5 marks]
Answer
- $x=\dfrac54$
Method: Since $\tanh u=1/3$ , the double-angle identity gives $\cosh(2u)=\dfrac{1+\tanh^2u}{1-\tanh^2u}=\dfrac{1+1/9}{1-1/9}=5/4$ . Applying $\cosh$ to $\operatorname{arcosh}x=2u$ gives $x=\cosh(2u)=5/4$ , which is in the required domain $x\geq1$ .

CP-8.4 · Derive and use the logarithmic forms of the inverse hyperbolic functions.

$\operatorname{arsinh}x=\ln(x+\sqrt{x^2+1})$ for every real $x$ .
$\operatorname{arcosh}x=\ln(x+\sqrt{x^2-1})$ for $x\geq1$ , using the non-negative principal branch.
$\operatorname{artanh}x=\dfrac12\ln\left(\dfrac{1+x}{1-x}\right)$ for $|x|<1$ .
When deriving a logarithmic form, reject any algebraic root that would make $e^y\leq0$ , and retain the inverse function's domain restriction.

Tier 1 · Easy

1. Use a logarithmic form to evaluate $\operatorname{arsinh}(3/4)$ exactly.[2 marks]
Answer
- $\ln2$
Method: $\operatorname{arsinh}(3/4)=\ln(3/4+\sqrt{9/16+1})=\ln(3/4+5/4)=\ln2$ .

Tier 2 · Standard

1. Solve $\operatorname{arsinh}x=\ln5-\ln2$ , giving $x$ exactly.[4 marks]
Answer
- $x=\dfrac{21}{20}$
Method: The right side is $\ln(5/2)$ . Applying $\sinh$ gives $x=\frac12(5/2-2/5)=\frac12(21/10)=21/20$ .

Tier 3 · Hard

1. Using logarithmic forms, prove that $\operatorname{artanh}\left(\dfrac{x}{\sqrt{1+x^2}}\right)=\operatorname{arsinh}x$ for every real $x$ .[6 marks]
Answer
- $\operatorname{artanh}\left(\dfrac{x}{\sqrt{1+x^2}}\right)=\ln(x+\sqrt{1+x^2})=\operatorname{arsinh}x$ .
Method: Put $s=\sqrt{1+x^2}$ . Since $|x|<s$ , the inverse-tanh input is valid. Its logarithmic form is $\frac12\ln((s+x)/(s-x))$ . But $(s-x)(s+x)=s^2-x^2=1$ , so $(s+x)/(s-x)=(s+x)^2$ . As $s+x>0$ , this becomes $\ln(s+x)$ , which is the logarithmic form of $\operatorname{arsinh}x$ .

CP-8.5 · Integrate functions of the form (x^2 + a^2)^(-1/2) and (x^2 - a^2)^(-1/2) and be able to choose substitutions to integrate associated functions.

For $\sqrt{x^2+a^2}$ , use $x=a\sinh u$ so that $\sqrt{x^2+a^2}=a\cosh u$ .
For $\sqrt{x^2-a^2}$ on $x\geq a>0$ , use $x=a\cosh u$ so that $\sqrt{x^2-a^2}=a\sinh u$ .
The standard antiderivatives are $\operatorname{arsinh}(x/a)+C$ for $(x^2+a^2)^{-1/2}$ and $\operatorname{arcosh}(x/a)+C$ for $(x^2-a^2)^{-1/2}$ on the stated branch.
Transform $dx$ and the limits as well as the radical; omitting one of these is the usual source of an incorrect scale factor.

Tier 1 · Easy

1. Find $\displaystyle\int\dfrac{1}{\sqrt{x^2+16}}\,dx$ .[2 marks]
Answer
- $\operatorname{arsinh}(x/4)+C$ , equivalently $\ln(x+\sqrt{x^2+16})+C$ .
Method: Use the standard form with $a=4$ : the antiderivative is $\operatorname{arsinh}(x/4)+C$ . Its logarithmic form differs from $\ln(x+\sqrt{x^2+16})$ only by the constant $-\ln4$ .

Tier 2 · Standard

1. Using $x=3\sinh u$ , evaluate $\displaystyle\int_0^4\sqrt{x^2+9}\,dx$ exactly.[6 marks]
Answer
- $10+\dfrac92\ln3$
Method: With $x=3\sinh u$ , $dx=3\cosh u\,du$ and $\sqrt{x^2+9}=3\cosh u$ . The limits are $0$ and $\operatorname{arsinh}(4/3)=\ln3$ . Hence the integral is $9\int_0^{\ln3}\cosh^2u\,du=9[u/2+\sinh(2u)/4]_0^{\ln3}$ . Since $\sinh(2\ln3)=40/9$ , the value is $(9/2)\ln3+10$ .

Tier 3 · Hard

1. Use a hyperbolic substitution to evaluate $\displaystyle\int_3^5\dfrac{2x^2-9}{\sqrt{x^2-9}}\,dx$ exactly.[6 marks]
Answer
- $20$
Method: Let $x=3\cosh u$ . Then $dx=3\sinh u\,du$ , $\sqrt{x^2-9}=3\sinh u$ , and the limits are $u=0$ and $u=\operatorname{arcosh}(5/3)=\ln3$ . The integrand becomes $18\cosh^2u-9=9\cosh(2u)$ . Therefore the integral is $9\int_0^{\ln3}\cosh(2u)\,du=(9/2)\sinh(2\ln3)=(9/2)(40/9)=20$ .