Question

Find the equation of the line that is normal to the graph of \(f(x)=x \cdot 5^{x}\) at the point where \(x=1\). Graph both the function and the normal line.

Short answer

The normal line equation is \( y = -\frac{1}{5 + 5 \ln(5)}x + \frac{1}{5 + 5 \ln(5)} + 5 \).

Step-by-step solution

Find the derivative of f(x)

To find the slope of the tangent at the point, we need to find the first derivative of the function. The given function is \( f(x) = x \cdot 5^x \). We use the product rule: \( f'(x) = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx} \), where \( u = x \) and \( v = 5^x \). The derivative of \( u = x \) is \( \frac{du}{dx} = 1 \) and the derivative of \( v = 5^x \) is \( \frac{dv}{dx} = 5^x \ln(5) \). Applying the product rule: \( f'(x) = 1 \cdot 5^x + x \cdot 5^x \ln(5) = 5^x (1 + x \ln(5)) \).

Evaluate the derivative at x = 1

Substitute \( x = 1 \) into the derivative \( f'(x) = 5^x(1 + x \ln(5)) \): \( f'(1) = 5^1 (1 + 1 \cdot \ln(5)) = 5(1 + \ln(5)) = 5 + 5 \ln(5) \).

Calculate the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line found in Step 2. Therefore, the slope of the normal line is:\(-\frac{1}{5 + 5 \ln(5)}\).

Find the coordinates where x = 1

The function value at \( x = 1 \) is \( f(1) = 1 \cdot 5^1 = 5 \). Thus, the coordinates of the point are \((1, 5)\).

Write the equation of the normal line

Using the point-slope form of the line equation, \( y - y_1 = m (x - x_1) \), where \( (x_1, y_1) = (1, 5) \) and \( m = -\frac{1}{5 + 5 \ln(5)} \):\( y - 5 = -\frac{1}{5 + 5 \ln(5)} (x - 1) \).Simplifying gives:\( y = -\frac{1}{5 + 5 \ln(5)}x + \frac{1}{5 + 5 \ln(5)} + 5 \).

Graph the function and the normal line

Graph the function \( f(x) = x \cdot 5^x \), showing its exponential growth. Then, plot the normal line using the equation found in Step 5, ensuring both intersect at the point \((1, 5)\).

Key concepts

Derivative

In calculus, a derivative is a measure of how a function changes as its input changes. Derivatives pave the way for understanding the behavior of functions by showing the rate of change. Think of it as telling us the slope of the tangent line to a curve at any given point on the graph. For the function \( f(x) = x \cdot 5^x \), the derivative is calculated to predict how the function's value changes when \( x \) changes.

Finding derivatives, especially for functions like products of simpler functions, uses specific rules such as the product rule. The derivative helps us determine key features of the graph, like the slope of the tangent line, which is crucial in solving the exercise here.

Normal line

A normal line to a curve at a given point is a line that is perpendicular to the tangent at that point. While the tangent line touches the curve and runs parallel to the curve's slope, the normal line stands at a right angle.

Finding the normal line is straightforward. We take the negative reciprocal of the tangent's slope to find the normal line's slope. In our exercise, the normal line's slope becomes \(-\frac{1}{5 + 5 \ln(5)}\).

This line is significant in many applications, such as optics and architecture, where perpendicular angles are analyzed.

Product rule

The product rule is a fundamental principle in calculus for differentiating products of two functions. If you have a function \( f(x) = u(x) \cdot v(x) \), then the derivative \( f'(x) \) is given by \( \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx} \).

In layman's terms, it's about taking the derivative of one function while keeping the other fixed, and then swapping. You sum these two results together.

For our function \( f(x) = x \cdot 5^x \), we identified \( u = x \) and \( v = 5^x \). Each component was differentiated separately: \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = 5^x \ln(5) \). The product rule then combines them to yield \( f'(x) = 5^x (1 + x \ln(5)) \). This approach is essential for any situation where functions multiply together.

Tangent line

The tangent line to a function at a given point is a straight line that just "touches" the curve at that point and shares the curve’s instantaneous direction at that spot. It provides an approximation of the curve near that specific point.

Calculating the tangent line involves finding the derivative of the function, as it gives us the slope of the tangent. At the point \( x = 1 \) on our function, the slope of the tangent line, \( 5^x (1 + x \ln(5)) \), becomes \( 5 + 5 \ln(5) \). This value tells us the direction in which the curve is moving and how steep it is at that point.

Tangent lines have wide applications, from physics to engineering, as they can help predict how a system will behave in the small neighborhood of a point.