Question

To determine the point of the curve\(y = coshx\)when the curve has slope \(1\).

Short answer

The slope of the curve is 1 at the point\(x = \ln (1 + \sqrt 2 )\).

Step-by-step solution

Given curve

The curve is \(y = \cosh x\).

The Concept of identities

The identity used:\(\frac{d}{{dx}}(coshx) = sinhx\)

The solution of quadratic equation is,\(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\).

Use the identities for solution

The slope of the tangent to the curve \(y = \cosh x\) is computed as shown below.

\(\begin{array}{c}\frac{{dy}}{{dx}} = \frac{d}{{dx}}(\cosh x)\\ = \sinh x\end{array}\)

The slope of the tangent to curve is, \(m = \sinh x\).

Suppose the slope of the tangent at \((x,y)\) is \(1\).

\(\begin{array}{c}\sinh x = 1\frac{{{e^x} - {e^{ - x}}}}{2}\\ = 1\frac{{{e^{2x}} - 1}}{{2{e^2}}}\\ = 1{\left( {{e^x}} \right)^2} - 2{e^x} - 1\\ = 0{\rm{ }}\end{array}\)

Let, \(x = {e^x},{x^2} - 2x - 1 = 0\).

Substitute the values

Substitute \(a = 1,b = - 2\) and \(c = - 1\) in \(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\).

\(\begin{array}{c}x = \frac{{2 \pm \sqrt {4 - 4(1)} }}{{2(1)}}\\ = \frac{{2 \pm \sqrt 8 }}{2}\\ = \frac{{2 \pm 2\sqrt 2 }}{2}\\ = 1 \pm \sqrt 2 \end{array}\)

Thus, \(x = 1 + \sqrt 2 \) or \(x = 1 - \sqrt 2 \).

Note that, the value of \({e^x}\) is positive.

Therefore, the value of \(x\) is \(1 + \sqrt 2 \).

Substitute, \(x = {e^x}\).

\(\begin{array}{c}{e^x} = 1 + \sqrt 2 x\\ = \ln (1 + \sqrt 2 )\end{array}\)

Therefore, the slope of the curve is \(1\) at the point \(x = \ln (1 + \sqrt 2 )\).