Question

Find an equation of the tangent line to the graph of the function at the given point. Then use a graphing utility to graph the function and the tangent line in the same viewing window. $$ f(t)=\left(t^{2}-9\right) \sqrt{t+2} \quad(-1,-8) $$

Short answer

The equation of the tangent line to the given function at the point (-1,-8) is \( y = -2x - 6 \)

Step-by-step solution

Identifying the Function and the Point

The given function is \( f(t) = (t^{2}-9) \sqrt{t+2} \) and the point of tangency is (-1,-8).

Calculating the Derivative

The function \( f(t) = (t^{2}-9) \sqrt{t+2} \) is a product of two functions, hence, the product rule of differentiation is used. First, let's rewrite \( \sqrt{t+2} \) as \( (t+2)^{0.5} \). The derivative of the function \( f'(t) \) is calculated as: \( f'(t) = 2t(t+2)^{0.5} + 0.5(t^{2}-9)(t+2)^{-0.5} \). Simplifying this we get \( f'(t) = 2t(t+2)^{0.5} + 0.5(t^{2}-9)(t+2)^{-0.5} \).

Calculating the Tangent Slope

Substituting the x-value of the point of tangency (-1) into the derivative, we get \( f'(-1) = 2*(-1)*(-1+2)^{0.5} + 0.5*(-1^{2}-9)(-1+2)^{-0.5} = -2 \). So, the slope of the tangent line is -2.

Derivation of the Tangent Line Equation

Using the point-slope form of a line equation, \( y - y1 = m(x - x1) \), where m is the slope and (x1, y1) is the point of tangency, we substitute these values: \( y - (-8) = -2(x - (-1)) \). Simplifying this we get the equation of the tangent line as \( y = -2x - 6 \)

Key concepts

Product Rule of Differentiation

When facing the task of differentiating the product of two functions, the product rule is the tool you need. It states that to find the derivative of a product of two functions, we differentiate the first function and multiply it by the second function, add the first function and multiply it by the derivative of the second function.

Consider our function from the exercise, which is a product of a polynomial, \(t^2-9\), and a root function, \(\sqrt{t+2}\). Applying the product rule, we differentiate \(t^2-9\) to get \(2t\) while keeping \(\sqrt{t+2}\) unchanged, then add \(t^2-9\) times the derivative of \(\sqrt{t+2}\), which is \(\frac{1}{2\sqrt{t+2}}\). A simplified form of the product rule is the powerful shortcut: \(f(t)g(t))' = f'(t)g(t) + f(t)g'(t)\).

By mastering the product rule, you'll be well-equipped to tackle more complex derivatives such as the one in our exercise.

Slope of a Tangent

The slope of a tangent is a measure of how steep the line is that just touches, or 'tangents', a curve at a particular point. In calculus, this is represented by the derivative of the function at that point. The derivative gives us the rate of change of the function's y-value with respect to its x-value, i.e., the slope of the curve.

For the given function in our exercise, we found the derivative using the product rule. To find the slope of the tangent at the given point, \(t=-1\), we simply substitute \(t\) with \-1 into the derivative. The result is \(f'(-1)\), which represents the slope of the tangent line at that specific point on the function. Knowing this slope allows us to pinpoint the exact direction and steepness of our tangent line.

Point-Slope Form

The point-slope form of a line, represented as \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \(\left(x_1, y_1\right)\) is a point the line passes through, is extremely helpful when writing the equation of a line if you know these two pieces of information.

In the context of the exercise, we have the slope \(m\) from the derivative at the point \(\left(-1, -8\right)\) and the point itself. Plugging these values into the point-slope form gives us the precise equation for the tangent line. Remember, this form is a straightforward method for writing an equation of a line, especially when working with tangents to curves due to already having a specific point and a slope.

Graphing Functions

Graphing functions is not just about drawing curves on a plane; it's also a valuable way to visually represent mathematical concepts and understand the behavior of functions. When graphing the original function and its tangent line, you can see how the tangent line touches the curve at the given point and the way the slope of the tangent line represents the rate of change of the function at that point.

Using graphing utilities or software, you can enter the equation of the function and the equation of the tangent line to see both the abstract relationships and their concrete representations. This visual aid is especially helpful in confirming the accuracy of arithmetic and understanding the geometric interpretation of derivatives, tangents, and other calculus concepts.