Question

Tangent lines and exponentials. Assume \(b\) is given with \(b>0\) and \(b \neq 1 .\) Find the \(y\) -coordinate of the point on the curve \(y=b^{x}\) at which the tangent line passes through the origin. (Source: The College Mathematics Journal, 28, Mar 1997).

Short answer

The y-coordinate of the point is \(b^{\frac{1}{-\ln(b)}}\).

Step-by-step solution

Differentiate the function y = \(b^x\)

To find the derivative of y = \(b^x\), we will use the following rule: If y = \(b^x\), then y' = \(\ln(b) b^x\). In this case, our function is y = \(b^x\). So, the derivative will be: y' = \(\ln(b) b^x\).

Find the point on the curve y = \(b^x\)

Let's call the point that we are looking for as P(x, \(b^x\)), where x is the corresponding x-coordinate of the point.

Use the point-slope equation to find the equation of the tangent line

The point-slope equation is given by: y - y1 = m(x - x1) In our case, the point P(x, \(b^x\)) is on the tangent line, and the slope of the tangent line at this point is given by the derivative y' = \(\ln(b) b^x\). So, we can write the equation of the tangent line as: y - \(b^x\) = \(\ln(b) b^x\) (x - x).

Find the y-coordinate of the point where the tangent line passes through the origin

Since the tangent line passes through the origin, we know that the values x = 0 and y = 0 are in the point-slope equation. By plugging in these values, we get: 0 - \(b^x\) = \(\ln(b) b^x\) (0 - x). Now, let's simplify this equation, \(b^x\) = - x\(\ln(b) b^x\) Divide both sides by \(b^x\): 1 = - x\(\ln(b)\) Now, divide both sides by -\(\ln(b)\): \(\frac{1}{-\ln(b)}\) = x Now, we found the x-coordinate of the point. So, let's plug this value back into the equation y = \(b^x\) to find the y-coordinate: y = \(b^{\frac{1}{-\ln(b)}}\) Now, we have the y-coordinate of the point on the curve y = \(b^x\) at which the tangent line passes through the origin.

Key concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it. This unique point of intersection is crucial because it provides the slope of the curve at that specific point. The slope of this tangent line reflects the direction the curve is heading at this point.
To find the tangent line to a curve, like the one given in the exercise, you need to determine:

• The point of tangency on the curve, defined as a point \(P(x, y)\).
• The slope of the tangent line at that point, which involves differentiation.

For the curve \(y = b^x\), the task is to find such a tangent line that also passes through the origin \((0, 0)\). This involves taking the derivative of the function \(y = b^x\) to get the slope at any point \(x\). Once we have the slope, we can work through the point-slope form of a linear equation to find the exact equation of the tangent line.

Exponential Functions

Exponential functions take the form \(y = b^x\), where \(b\) is a constant greater than zero, but not equal to one. These functions are fundamental in many fields, from science to engineering, due to their growth properties.
Here's a breakdown of exponential functions:

• The base \(b\) determines the growth rate. If \(b > 1\), the function represents exponential growth, whereas if \(0 < b < 1\), it represents exponential decay.
• As \(x\) increases, \(y = b^x\) grows rapidly for \(b > 1\), often used to model phenomena like population growth and radioactive decay.
• The graph of this function is always a curve, steadily increasing or decreasing without turning back.

In the given problem, understanding the behavior of \(y = b^x\) aids in visualizing where the tangent line would pass through the origin. This requires manipulating the function to find specific points that satisfy the tangent line's conditions.

Derivative

The derivative of a function at a point gives us the slope of the tangent line to the curve at that point. It's a fundamental concept in calculus that helps in understanding the rate at which a function changes.
Calculating the derivative:

• For the exponential function \(y = b^x\), the derivative is \(y' = \ln(b) b^x\).
• This result comes from the rule for derivatives of exponential functions, typically involving the natural logarithm \(\ln\) of the base \(b\).

Once the derivative is known, it can be used to solve problems involving tangent lines. It allows us to develop an equation for the tangent line by substituting the derivative into the point-slope formula \(y - y_1 = m(x - x_1)\). This way, the derivative directly contributes to finding the slope \(m\) at any point \(P(x, b^x)\) on the curve.