Problem 2 - Calculus an Applied Approach

Question

Find the slope of the tangent line to \(y=x^{n}\) at the point \((1,1) .\) $$ \text { (a) } y=x^{3 / 2} \quad \text { (b) } y=x^{3} $$

Step-by-Step Solution

Answer

The slope of the tangent line to \(y=x^{3/2}\) at \((1,1)\) is \(\frac{3}{2}\) and to \(y=x^{3}\) at \((1,1)\) is \(3\).

Step 1: Find the derivative of the functions

For the first function \(y=x^{3/2}\), the derivative using the power rule is \(\frac{3}{2}x^{3/2-1} = \frac{3}{2}x^{1/2}\). For second function \(y=x^{3}\), the derivative using the power rule is \(3x^{3-1} = 3x^{2}\).

Step 2: Evaluate the derivative at the given point

At the point, \((1,1)\), for the first function \(y=x^{3/2}\), the slope of the tangent is \(\frac{3}{2}*(1)^{1/2} = \frac{3}{2}\). For the second function \(y=x^{3}\), the slope of the tangent is \(3*(1)^{2} = 3\).

Step 3: State the solutions

The slope of the tangent line to the function \(y=x^{3/2}\) at \((1,1)\) is \(\frac{3}{2}\) and to the function \(y=x^{3}\) at \((1,1)\) is \(3\). Thus, these are the answers.

Key Concepts

Derivative

Understanding the concept of the derivative is critical in calculus. It's a measure of how a function changes as its input changes. In essence, the derivative represents the rate of change or the slope of the curve of a function at a particular point. When you calculate the derivative of a function, you're finding the slope of the tangent line to that function's graph at any point.

It's this tangent line that gives you the exact slope at a single point, differentiating it from an average slope calculated over an interval. For practical applications, derivatives are used to determine the maximum and minimum values of a function, and they're essential in fields like physics, where they express rates like velocity and acceleration.

Power Rule

The power rule is a quick method to differentiate functions of the form \(y=x^n\), where \(n\) is any real number. According to this rule, the derivative of \(y=x^n\) is \(dy/dx = nx^{n-1}\). This method is incredibly useful because it simplifies the differentiation process, especially when dealing with polynomial functions.

For example, when applying the power rule to the function \(y=x^{3/2}\), we multiply the exponent by the coefficient (which is 1 in this case, though it's often omitted) and then subtract one from the exponent to get \(dy/dx = (3/2)x^{1/2}\), which is the slope of the tangent line at any point on the curve.

Tangent Line Calculus

In calculus, a tangent line represents a line that just touches a curve at a specific point, sharing the same instantaneous rate of change or derivative at that point. This is what makes tangent lines so unique - they are a local linear approximation of the curve at a particular point.

To find the equation of a tangent line, you'll need two essential pieces of information: the point of tangency and the slope of the tangent line at that point. By finding the derivative of the function, and evaluating it at the given point, you get the slope. Then, using the point-slope form of a line equation, you can write out the equation for the tangent line. As an illustration, for the function \(y=x^3\), the derivative at point \((1,1)\) gives us a slope of 3, leading to the tangent line \(y-1=3(x-1)\), which represents the tangent to the curve at \((1,1)\).