Problem 32 - Applied Calculus

Question

Find the equation of the tangent line to \(f(x)=10 e^{-0.2 x}\) at \(x=4\).

Step-by-Step Solution

Answer

The tangent line's equation is \(y = -0.8986x + 7.0874\).

Step 1: Find the function derivative

To determine the equation of the tangent line, we need the slope at the point of tangency, which is given by the derivative of the function. The function is \(f(x) = 10e^{-0.2x}\). To find its derivative, we'll use the chain rule. The derivative of \(e^{u}\) is \(e^{u}\cdot u'\). For our function, \(u = -0.2x\) and \(u' = -0.2\). Therefore, the derivative is:\[f'(x) = 10 \cdot (-0.2)e^{-0.2x} = -2e^{-0.2x}\].

Step 2: Calculate the slope at x=4

Now, substitute \(x=4\) into the derivative to find the slope of the tangent line at that point.\[f'(4) = -2e^{-0.2 \times 4} = -2e^{-0.8}\].Calculate \(e^{-0.8}\) to find the exact slope value. Using a calculator, \(e^{-0.8} \approx 0.4493\). Therefore, \[f'(4) \approx -2 \times 0.4493 = -0.8986\].

Step 3: Find the y-coordinate at x=4

Plug \(x=4\) back into the original function to find the y-coordinate of the point of tangency:\[f(4) = 10e^{-0.2 \times 4} = 10e^{-0.8}\].Using a calculator, \(e^{-0.8} \approx 0.4493\), so \[f(4) \approx 10 \times 0.4493 = 4.493\].

Step 4: Write the equation of the tangent line

The equation of a line in point-slope form is given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point of tangency and \(m\) is the slope. We have \(x_1 = 4\), \(y_1 = 4.493\), and \(m = -0.8986\). Substitute these values into the equation:\[y - 4.493 = -0.8986(x - 4)\].Simplify and convert to slope-intercept form:\[y = -0.8986x + 7.0874\].

Key Concepts

Derivative

In calculus, the derivative of a function tells us how the function's output changes with respect to changes in its input. It's essentially the slope of the function at any given point. To find the derivative of a function, we often apply rules such as the power rule or the chain rule. In the given exercise, we found the derivative of the function \( f(x) = 10e^{-0.2x} \). By applying the chain rule, which helps us differentiate composite functions, we determined that the derivative is \( f'(x) = -2e^{-0.2x} \). Here, the chain rule was necessary because the exponent \(-0.2x\) is a linear function of \(x\). This derivative indicates how steeply the function is increasing or decreasing at any point \(x\).

• The chain rule simplifies the differentiation of composite functions.
• Derivative calculation provides the slope of the tangent line at a point.

Tangent Line

A tangent line to a curve at a given point is a straight line that just “touches” the curve at that point. It shares the same slope as the curve at the point of tangency. The concept of the tangent line is majorly beneficial to understanding the instantaneous rate of change. In our example, the tangent line at \( x=4 \) gives information about how the function \( f(x) \) behaves around that point. By using the derivative, we found the slope of this tangent line, \( m = -0.8986 \). The point-slope form of a line helps us to write the equation of the tangent line using this slope and the point of tangency: \( (x_1, y_1) = (4, 4.493) \).

• Tangent lines provide local linear approximations of functions.
• They demonstrate the direction and rate of change of functions.

Chain Rule

Understanding the chain rule is crucial when working with derivatives, particularly with composite functions. The chain rule states that to differentiate a composite function \( f(g(x)) \), one must multiply the derivative of the outer function by the derivative of the inner function. In this exercise, the function involved was \( f(x) = 10e^{-0.2x} \), where \( e^{-0.2x} \) is a composite function. We considered \( u = -0.2x \), giving us \( u' = -0.2 \). By applying the chain rule, we determined the overall derivative as \( f'(x) = -2e^{-0.2x} \), where \( -2 \) results from multiplying the inner derivative by the constant factor 10.

• The chain rule simplifies differentiation of nested functions.
• It is essential in calculus to correctly compute derivatives of complex expressions.

Exponential Function

Exponential functions are functions of the form \( a^x \), where \( a \) is a positive constant. They are unique in that the rate of change is directly proportional to the value of the function itself. In calculus, the natural exponential function \( e^x \) is particularly important due to its derivative, \( (e^x)'=e^x \). In our problem, the function was \( f(x) = 10e^{-0.2x} \), showcasing the decay of the function, with \( -0.2 \) indicating the rate of decay. Exponential functions are used to model many real-world phenomena, such as population growth and radioactive decay.

• Exponential functions have a constant proportional growth rate.
• The natural base \( e \) is central in continuous growth models.