Problem 4 - Calculus an Applied Approach

Question

Find the slope of the tangent line to \(y=x^{n}\) at the point \((1,1) .\) $$ \text { (a) } y=x^{-1 / 2} \quad \text { (b) } y=x^{-2} $$

Step-by-Step Solution

Answer

The slope of the tangent line to the first function \( y = x^{-1 / 2} \) at the point (1,1) is -0.5 and for the second function \( y = x^{-2} \) at the same point is -2.

Step 1: Differentiate The First Function

Differentiate \(y = x^{-1/2} \). By applying the power rule of differentiation, i.e. \( y' = n∗x^{n−1} \), we get: \(y' = -\frac{1}{2} x^{-3/2} = -\frac{1}{2\sqrt{x^3}}\).

Step 2: Find the slope of the first function at the point (1,1)

Substitute \( x = 1 \) into the derivative to find the slope at that point. When \( x = 1 \), the derivative \(y'\) is equal to -0.5.

Step 3: Differentiate The Second Function

Differentiate \( y = x^{-2} \). By applying the power rule, we get: \(y' = -2x^{-3} = -2/x^3 \).

Step 4: Find the slope of the second function at the point (1,1)

Again, substitute \( x = 1 \) into the derivative obtained in step 3 to find the slope at that point. So, for \( x = 1 \), \(y'\) is -2.

Key Concepts

Power Rule of Differentiation

The power rule is a fundamental concept in calculus, specifically in the field of differentiation. It is a technique that allows us to find the derivative of a function that is a power of x. The derivative represents the rate of change of a function, which is essential in understanding how a function behaves.
According to the power rule, if we have a function of the form y = x^n, where n is a real number, the derivative y' of the function with respect to x is given by:
y' = nx^(n−1).
In the context of the original exercise, we have two functions, y = x^(-1/2) and y = x^(-2). Applying the power rule, the derivatives of these functions are y' = -1/2 x^(-3/2) and y' = -2 x^(-3), respectively. This rule simplifies the process of differentiation by providing a direct formula to find the slope of the curve at any point x. The importance of the power rule cannot be overstated – it is one of the first techniques taught for finding derivatives and is widely used across different fields of science and engineering.

Derivative of Exponential Function

While the original exercise doesn't explicitly deal with exponential functions, understanding the derivative of an exponential function is invaluable in calculus. An exponential function is of the form f(x) = a^x, where a is a constant, and it has a unique property that its rate of change is proportional to its value.
For the case where the base a is the special number e (Euler's number, approximately 2.71828), the function f(x) = e^x has a fascinating characteristic: its derivative is f'(x) = e^x. That means the slope of the tangent line to the curve at any point is equal to the value of the function at that point.
The knowledge of exponential functions and their derivatives is crucial when working with growth and decay processes in various disciplines, such as biology, finance, and physics. The exponential function models processes where changes occur at rates proportional to current values, making the differential calculus a powerful tool in analyzing such functions.

Tangent Line Calculus

In calculus, a tangent line is a line that touches a curve at a single point without crossing over the curve. It is representative of the instantaneous rate of change of the function at that specific point, which is another way of saying it represents the derivative of the function at a point.
The slope of the tangent line at any given point on a function can be found by taking the derivative of that function and evaluating it at the point of interest. In the exercise provided, we find the slopes of the tangent lines to the functions y = x^{-1/2} and y = x^{-2} at the point (1,1). Through the power of differentiation, we deduced that the slopes are -0.5 and -2, respectively.
Understanding the concept of a tangent line is vital for solving many real-world problems, from determining the best angle to kick a soccer ball to optimizing the design of a car's suspension system. By mastering this concept, students can gain a deeper insight into not just how, but why different functions behave the way they do.