Tangent Slope at Point y=\cosh x = 1

Thread starter bard
Start date Dec 3, 2003
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Hyperbolic

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The discussion centers on finding the point on the curve y = cosh x where the tangent has a slope of 1. The user initially expresses uncertainty about how to approach the problem but later derives the relationship between the slope and the derivative, leading to the equation 1 = sinh x * dy/dx. By utilizing the definition of sinh, they ultimately determine that the solution occurs at x = ln(1 + √2). The conversation highlights the importance of understanding hyperbolic functions and their derivatives in solving such problems.

Dec 3, 2003

bard

at what point on the curve y=\cosh x does the tangent have slope 1

I have no idea how to approach this problem

my work

1=\sinh x\frac{dy}{dx}

\frac{1}{sinh x}=\frac{dy}{dx}

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Dec 3, 2003

NateTG

Science Advisor

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It would probably help to you use the definitions.

For example:
\sinh{x}= \frac{e^{x}-e^{-x}}{2}

Dec 3, 2003

bard

yah I found the answer as x=ln[1+sqt2]----thanks for your help

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