Question

Definitions of hyperbolic sine and cosine Complete the following steps to prove that when the x- and y-coordinates of a point on the hyperbola x² - y² = 1 are defined as cosh t and sinh t, respectively, where t is twice the area of the shaded region in the figure, x and y can be expressed as
x = cosh t = (eᵗ + e⁻ᵗ) / 2 and y = sinh t = (eᵗ - e⁻ᵗ) / 2.

b. In Chapter 8, the formula for the integral in part (a) is derived:
∫ √(z² − 1) dz = (z/2)√(z² − 1) − (1/2) ln|z + √(z² − 1)| + C.
Evaluate this integral on the interval [1, x], explain why the absolute value can be dropped, and combine the result with part (a) to show that:
t = ln(x + √(x² − 1)).

Accepted Answer

Start by expressing the parameter $$ t  as $$twice the shaded area, which is given by the formula: $$ t = 2 \left( \frac{1}{2} x y - \int_1^x \sqrt{z^2 - 1} \, dz \right) . $$This represents the area of the triangle minus the area under the curve from 1 to $$ x . $$Recognize that the point $$ P(x, y) $$ lies on the hyperbola $$ x^2 - y^2 = 1 $$, and the coordinates are defined as $$ x = \cosh t $$ and $$ y = \sinh t . $$This means $$ y = \sqrt{x^2 - 1} $$ because $$ y^2 = x^2 - 1 . $$Use the integral formula provided: $$ \int \sqrt{z^2 - 1} \, dz = \frac{z}{2} \sqrt{z^2 - 1} - \frac{1}{2} \ln|z + \sqrt{z^2 - 1}| + C . $$Evaluate this definite integral from 1 to $$ x  by $$substituting the limits into the antiderivative. Explain why the absolute value in the logarithm can be dropped: since $$ x \geq 1 $$ and $$ \sqrt{x^2 - 1} \geq 0 $$, the expression inside the logarithm, $$ x + \sqrt{x^2 - 1} $$, is always positive, so $$ |x + \sqrt{x^2 - 1}| = x + \sqrt{x^2 - 1} . $$Combine the evaluated integral with the expression for $$ t $$ from step 1, simplify the terms, and show that $$ t = \ln(x + \sqrt{x^2 - 1}) . $$This links the parameter $$ t  to $$the coordinates on the hyperbola and completes the proof.

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Key Concepts

Hyperbolic Functions (sinh and cosh)

Area Interpretation and Parametrization of the Hyperbola

Integral of √(z² - 1) and Logarithmic Expression