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 on the curve y = b^x at which the tangent line passes through the origin is y_0 = b^(1/ln(b)).

Step-by-step solution

Find the derivative of the given function

To find the slope of the tangent line at any point on the curve, we first need to find the derivative of the function \(y = b^x\). Using the exponent rule, it can be found as $$\frac{dy}{dx} = b^x \cdot \ln{b}.$$

Find the equation of the tangent line

Next, we find the equation of the tangent line at a point \((x_0, y_0)\) on the curve. Using the point-slope form, we get the equation of the tangent line as $$y - y_0 = (b^{x_0} \cdot \ln{b})(x - x_0).$$

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

Since the tangent line passes through the origin (0,0), we can substitute \((x, y) = (0,0)\) into the tangent line equation and solve for \(y_0\). We get $$0 - y_0 = (b^{x_0} \cdot \ln{b})(0 - x_0).$$ Now, we know that \(y_0 = b^{x_0}\), so we can write the equation as $$0 - b^{x_0} = (b^{x_0} \cdot \ln{b})(0 - x_0).$$ Divide both sides by \(-b^{x_0}\): $$1 = \ln{b} \cdot x_0.$$ Finally, solve for \(x_0\): $$x_0 = \frac{1}{\ln{b}}.$$

Calculate the \(y\)-coordinate using the found \(x_0\) value

Now that we have the \(x\)-coordinate of the point where the tangent line passes through the origin, we can substitute it back into the equation of our curve (\(y = b^x\)) to find the corresponding \(y\)-coordinate: $$y_0 = b^{\frac{1}{\ln{b}}}.$$ Thus, the \(y\)-coordinate of the point on the curve \(y = b^x\) at which the tangent line passes through the origin is \(b^{\frac{1}{\ln{b}}}\)

Key concepts

Derivative of Exponential Function

Understanding the derivative of an exponential function is key for tackling problems involving growth processes, decay, and finding slopes of tangent lines in calculus. The exponential function, typically expressed as y = b^x, where b is a positive constant, has a unique property – its derivative is proportional to the function itself.

The derivative is given by the formula:
\[ \frac{dy}{dx} = b^x \cdot \ln{b}\]
Here, \(\ln{b}\) is the natural logarithm of b, which essentially provides the rate at which the function is changing at any point x. This concept is especially important in fields like economics, physics, and biology, where exponential relationships frequently occur.

When you're solving problems involving exponential functions, remember that knowing this derivative allows you to find tangent lines to the curve at any point, which is a major application in geometry and optimisation problems.

Point-Slope Form of a Line

If you need to write the equation of a line and you know a point it passes through and its slope, you'll turn to the point-slope form of a line. This is a straightforward method used in algebra and calculus, which can be articulated as:
\[ y - y_1 = m(x - x_1)\]
where \(m\) is the slope of the line and \((x_1, y_1)\) is any point on the line. It's often used because it can also help you shift to other forms of linear equations, like slope-intercept or standard form, with relative ease.

In the context of tangent lines to curves, the point-slope formula is particularly useful because it requires just the slope at a certain point—which you can find using differentiation—and the coordinates of the point itself. This creates a reliable bridge between the pure calculus of finding derivatives and the geometric interpretation of those derivatives as tangent lines.

Slope of a Tangent Line

The slope of a tangent line to a curve at a given point signifies the steepness of the curve at that exact location. In other words, the slope tells us how fast the value of y is changing relative to x at that point on the curve. To find it for an exponential function, we differentiate the function.

Mathematically speaking, if you have an exponential function y = b^x, the slope of the tangent line at any point x is determined by the function's derivative dy/dx. For the above mentioned function, at any point \(x_0\), the slope is \(b^{x_0} \cdot \ln{b}\).

Once you understand that the derivative represents the slope of the tangent line, you can explore many applications like rates of change in physics, calculating instantaneous velocity, and more, which are central topics in calculus.

Calculus Applications in Geometry

Calculus has profound applications in geometry, particularly when we explore the slopes of curves and surfaces. In geometry, the ability to determine tangent lines to various shapes is crucial for understanding their properties and describing their behavior.

For instance, when you find the derivative of an exponential function, you're not just performing an abstract calculation; you're actually finding the slope of a line that just 'touches' the curve at one particular point. This concept can be expanded to find normal lines (which are perpendicular to tangents), explore the curvature of an object, or even determine the shortest or longest possible distances in optimization problems.

In summary, calculus allows us to transition from static descriptions of shapes and figures to a dynamic understanding of how they change. This shift is essential in fields like physics, engineering, and computer graphics, where the real-world applications of geometry are both profound and practical.