91Ó°ÊÓ

When dealing with exponential functions, understanding their derivatives is crucial. Exponential functions generally take the form of \( e^{u(x)} \), where \( u(x) \) is a function of \( x \). The chain rule is often used to differentiate exponential functions. It states that \( \frac{d}{dx} [e^{u(x)}] = e^{u(x)} \cdot u'(x) \). This means you differentiate the exponent and multiply it by the original exponential function.In our exercise, we are given a function involving the sum of two exponential terms, \( y = \frac{a}{2} (e^{x/a} + e^{-x/a}) \). By applying the rule to each term, the derivative becomes \( \frac{dy}{dx} = \frac{1}{2} (e^{x/a} - e^{-x/a}) \). This result is achieved because the derivative of \( e^{x/a} \) is \( \frac{1}{a} e^{x/a} \) while \( e^{-x/a} \) yields \( -\frac{1}{a} e^{-x/a} \). Each is scaled by the factor of 1/2 from the original function.**Key Points:**- Differentiate each exponential term separately.- Use the chain rule for differing exponents.- Pay attention to multiplicative constants.

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola rather than a circle. They include \( \cosh(x) \) and \( \sinh(x) \), which are counterparts to cosine and sine.**Definitions:**- The hyperbolic cosine, \( \cosh(x) \), is defined as \( \frac{e^x + e^{-x}}{2} \).- The hyperbolic sine, \( \sinh(x) \), is \( \frac{e^x - e^{-x}}{2} \).In the exercise, we notice that the expression under the square root simplifies via these hyperbolic identities. We arrive at: \[ \sqrt{\cosh^2(x/a)} = \cosh(x/a) \]This simplification makes it easier to integrate later. Hyperbolic functions have many properties similar to trigonometric functions, such as analogous addition formulas and derivatives, making them powerful tools in calculus.**Key Insights:**- Recognize expressions that can simplify to hyperbolic forms.- Familiarity with hyperbolic identities can simplify integration tasks.

To find the arc length of a curve, we utilize the arc length formula: \[ \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]In our problem, we apply this to the function \( y = \frac{a}{2} (e^{x/a} + e^{-x/a}) \). The established derivative \( \frac{dy}{dx} = \frac{1}{2} (e^{x/a} - e^{-x/a}) \) was substituted into the arc length formula, leading to a new integrand.After simplifying the expression within the square root using hyperbolic functions, we get:\[ \cosh(x/a) \]Therefore, the integral that gives the distance traveled by the camera is:\[ \int_{-5}^{5} \cosh\left(\frac{x}{a}\right) \, dx \]Setting up the integral correctly is crucial as it directly affects the calculation of arc length. The boundaries \(-5\) to \(5\) ensure we're capturing the entire span of the curve defined in the problem.**Critical Steps to Remember:**- Use correct differential expression \( \frac{dy}{dx} \) in arc length formula.- Simplify wherever possible using identities.- Pay attention to integration limits for full path coverage.