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Assume a hanging cable has the shape 15 cosh ( x / 15 ) for −20 x 20 . Determine the length of the cable (in feet).

52.95 ft

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Key concepts

  • Hyperbolic functions are defined in terms of exponential functions.
  • Term-by-term differentiation yields differentiation formulas for the hyperbolic functions. These differentiation formulas give rise, in turn, to integration formulas.
  • With appropriate range restrictions, the hyperbolic functions all have inverses.
  • Implicit differentiation yields differentiation formulas for the inverse hyperbolic functions, which in turn give rise to integration formulas.
  • The most common physical applications of hyperbolic functions are calculations involving catenaries.

[T] Find expressions for cosh x + sinh x and cosh x sinh x . Use a calculator to graph these functions and ensure your expression is correct.

e x and e x

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From the definitions of cosh ( x ) and sinh ( x ) , find their antiderivatives.

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Show that cosh ( x ) and sinh ( x ) satisfy y = y .

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Use the quotient rule to verify that tanh ( x ) = sech 2 ( x ) .

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Derive cosh 2 ( x ) + sinh 2 ( x ) = cosh ( 2 x ) from the definition.

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Take the derivative of the previous expression to find an expression for sinh ( 2 x ) .

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Prove sinh ( x + y ) = sinh ( x ) cosh ( y ) + cosh ( x ) sinh ( y ) by changing the expression to exponentials.

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Take the derivative of the previous expression to find an expression for cosh ( x + y ) .

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For the following exercises, find the derivatives of the given functions and graph along with the function to ensure your answer is correct.

[T] cosh ( 3 x + 1 )

3 sinh ( 3 x + 1 )

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[T] 1 cosh ( x )

tanh ( x ) sech ( x )

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[T] cosh 2 ( x ) + sinh 2 ( x )

4 cosh ( x ) sinh ( x )

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[T] cosh 2 ( x ) sinh 2 ( x )

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[T] tanh ( x 2 + 1 )

x sech 2 ( x 2 + 1 ) x 2 + 1

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[T] 1 + tanh ( x ) 1 tanh ( x )

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[T] sinh 6 ( x )

6 sinh 5 ( x ) cosh ( x )

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[T] ln ( sech ( x ) + tanh ( x ) )

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For the following exercises, find the antiderivatives for the given functions.

cosh ( 2 x + 1 )

1 2 sinh ( 2 x + 1 ) + C

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x cosh ( x 2 )

1 2 sinh 2 ( x 2 ) + C

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cosh 2 ( x ) sinh ( x )

1 3 cosh 3 ( x ) + C

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tanh 2 ( x ) sech 2 ( x )

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sinh ( x ) 1 + cosh ( x )

ln ( 1 + cosh ( x ) ) + C

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cosh ( x ) + sinh ( x )

cosh ( x ) + sinh ( x ) + C

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( cosh ( x ) + sinh ( x ) ) n

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For the following exercises, find the derivatives for the functions.

tanh −1 ( 4 x )

4 1 16 x 2

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sinh −1 ( cosh ( x ) )

sinh ( x ) cosh 2 ( x ) + 1

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tanh −1 ( cos ( x ) )

csc ( x )

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ln ( tanh −1 ( x ) )

1 ( x 2 1 ) tanh −1 ( x )

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For the following exercises, find the antiderivatives for the functions.

d x a 2 x 2

1 a tanh −1 ( x a ) + C

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x d x x 2 + 1

x 2 + 1 + C

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e x e 2 x 1

cosh −1 ( e x ) + C

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For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation d v / d t = g v 2 .

Show that v ( t ) = g tanh ( g t ) satisfies this equation.

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Derive the previous expression for v ( t ) by integrating d v g v 2 = d t .

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[T] Estimate how far a body has fallen in 12 seconds by finding the area underneath the curve of v ( t ) .

37.30

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For the following exercises, use this scenario: A cable hanging under its own weight has a slope S = d y / d x that satisfies d S / d x = c 1 + S 2 . The constant c is the ratio of cable density to tension.

Show that S = sinh ( c x ) satisfies this equation.

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Integrate d y / d x = sinh ( c x ) to find the cable height y ( x ) if y ( 0 ) = 1 / c .

y = 1 c cosh ( c x )

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Sketch the cable and determine how far down it sags at x = 0 .

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Practice Key Terms 1

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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