Problem 61 - Calculus of a Single Variable

Question

Finding an Equation of a Tangent Line In Exercises \(55-62,\) find an equation of the tangent line to the graph of the function at the given point. $$ y=x^{2} e^{x}-2 x e^{x}+2 e^{x}, \quad(1, e) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the graph of the function \(y = \log_{3}x\) at the point (27,3) is \(y - 3 = \frac{1}{27 * \ln 3}(x - 27)\)

Step 1: Find the derivative of the given function

The derivative of \(y = \log_{3}x\) can be found using the derivative of natural logarithmic function and applying the base change rule. So, \(y' = \frac{1}{x\ln 3}\).

Step 2: Evaluate the derivative at the given point

To find the slope of the tangent line at the given point (27,3), substitute x value of the point into the derivative: \(y'(27) = \frac{1}{27* \ln 3}\). So, the slope of the tangent line at this point is \(m =\frac{1}{27* \ln 3}\).

Step 3: Use the point-slope form of an equation

Finally, using the point-slope form of an equation, which is \(y - y1 = m(x - x1)\), where (x1,y1) is the point (27,3), and m is the slope of the line, we derived in step 2. By substituting these values, we get \(y - 3 = \frac{1}{27 * \ln 3}* (x - 27)\).

Key Concepts

Derivative of Logarithmic Function

When you work with logarithmic functions in calculus, one of the crucial tasks is taking their derivatives. A logarithmic function with a base other than the natural base, like 3 in our exercise, can be transformed using the change of base formula: \( \log_b(x) = \frac{\ln(x)}{\ln(b)} \). This transformation helps when deriving the function.
The derivative of natural logarithmic function \( \ln(x) \) is \( \frac{1}{x} \). Applying this rule to the function \( y = \log_{3} x \), use the chain rule: the derivative is \( y' = \frac{\ln'(x)}{\ln 3} = \frac{1}{x \ln 3} \). This derivative tells how the log base 3 function changes at any point x.
Understanding these derivatives is a core skill in calculus, essential for solving problems related to growth rates and decay in various fields.

Slope of a Line

The slope of a line is a measure of its steepness and direction. It is a fundamental concept in calculus and coordinate geometry. In simple terms, the slope can be defined as rise over run. For a given linear equation or a tangent line, it is calculated as the ratio of the vertical change to the horizontal change between two points on the line.
When dealing with calculus problems, such as tangents on curves, the slope corresponds to the derivative of the function at a certain point. In our exercise, once the derivative of \( y = \log_{3}x \) was calculated, it was evaluated at the point \( x = 27 \) to get the slope, \( \frac{1}{27 \ln 3} \). This gives us insights into how sharply the function is increasing or decreasing at that particular point.

Point-Slope Form

The point-slope form is an equation that describes a line using a specific point and the slope of the line. It is incredibly useful, especially in calculus, when we want to determine the equation of a tangent line. The form is written as \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a known point on the line and \( m \) is the slope.
For the tangent line to \( y = \log_{3}x \) at point \( (27,3) \), the slope \( m = \frac{1}{27 \ln 3} \) was crucial. Substituting into the point-slope form gave us the tangent line equation: \( y - 3 = \frac{1}{27 \ln 3} (x - 27) \). This representation is vital in calculus for describing linear approximations of functions at specific points.

Calculus

Calculus is the branch of mathematics that deals with rates of change and the accumulation of quantities. One of its central ideas is differentiation, which allows us to find instantaneous rates of change, or derivatives. A derivative shows how a function changes as its input changes, and it becomes the foundation for finding anything from slopes of tangent lines to optimizations in real-world problems.

• Two major branches of calculus are differential calculus and integral calculus.
• Differential calculus primarily focuses on derivatives and their applications.
• Integral calculus centers on area accumulation and solving integrals.

Understanding key calculus concepts, like the derivative in this exercise, gives students the tools to analyze and solve various scientific and theoretical problems.