Question

(a) find an equation of the tangent line to the graph of \(f\) at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. $$ h(t)=\sin t+\frac{1}{2} e^{t} $$ $$ \left(\pi, \frac{1}{2} e^{\pi}\right) $$

Short answer

The equation of the tangent line to the graph of the function \(h(t) = \sin t + \frac{1}{2} e^{t}\) at the point \((\pi, \frac{1}{2} e^{\pi})\) is \(y - \frac{1}{2} e^{\pi} = (-1 + \frac{1}{2} e^{\pi})(x - \pi)\).

Step-by-step solution

Compute the Derivative of the Function

The derivative of the function \(h(t) = \sin t + \frac{1}{2} e^{t}\) is calculated using chain rule and the derivates of sin and exponential functions: \(h'(t) = \cos t + \frac{1}{2} e^{t}\).

Evaluate the Derivative at the Given Point

Next, substitute the given \(x\)-value (\(t = \pi\)) into the derivative to calculate the slope of the tangent line: \(h'(\pi) = \cos \pi + \frac{1}{2} e^{\pi} = -1 + \frac{1}{2} e^{\pi}\).

Write the Equation of the Tangent Line

Now that the slope \(m\) has been found, the equation of the tangent line can be calculated using the point-slope form. The point is \((\pi, \frac{1}{2} e^{\pi})\) and the slope is \(m = -1 + \frac{1}{2} e^{\pi}\). Substituting into the formula gets the equation \(y - \frac{1}{2} e^{\pi} = (-1 + \frac{1}{2} e^{\pi})(x - \pi)\).

Key concepts

Derivative of a Function

Understanding the derivative of a function is crucial when studying calculus as it represents the rate at which a function is changing at any given point. In essence, it's the slope of the tangent line to the function's graph at a specific point. To find a derivative, we apply rules such as the power rule, product rule, quotient rule, and, for more complex functions, the chain rule.

For the function in our exercise, we have a combination of a sine function and an exponential function. The derivative of the sine function, \(\sin t\), is \(\cos t\), while the derivative of the exponential function, \(e^{t}\), is simply \(e^{t}\). When these functions are combined, the derivatives are added together to get the overall derivative of the function.

Point-Slope Form

The point-slope form of a line's equation is specifically useful for when you have a point on the line and the slope of that line. The general formula is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \(\(x_1, y_1\)\) is the given point. This form makes it simple to write the equation of a line once these two pieces of information are known.

In our problem, with the slope \(m\) calculated from the derivative of the function at a specific point, and the point \(\(\pi, \frac{1}{2} e^{\pi}\)\), we can directly substitute these values into the point-slope formula to find the equation of the tangent line.

Graphing Utility

A graphing utility is an invaluable tool in calculus for visualizing functions and their characteristics including tangent lines. These utilities can plot the intricate shapes of functions based on their mathematical formulas and also provide features such as zooming, tracing, and evaluating derivatives at specific points.

By entering the function \(h(t)\) into a graphing utility, we can clearly see the behavior of the function and the exact point where the tangent line touches the graph. Adding the equation of the tangent line to the same utility allows for a visual confirmation of correct calculations and understanding the relationship between the function and the tangent.

Chain Rule

The chain rule is a formula for computing the derivative of the composition of two or more functions. This rule is essential when you're dealing with a function that is made up of other functions, commonly seen in calculus problems. In simpler terms, if a variable is being affected by multiple functions, the chain rule tells us how to find the rate of change of the entire system.

In the given exercise, although the function \(h(t)\) doesn't explicitly require the chain rule to calculate its derivative, understanding this rule is vital for advancing in calculus. It equips us to handle more complex functions where the application of this rule is necessary to find the derivative correctly.