Problem 59 - Calculus and its applications

Question

Find an equation for the tangent line to the graph of \(y=x \sqrt{2 x+3}\) at the point (3,9)

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = 4x - 3 \).

Step 1: Differentiate the Function

Let's find the derivative of the given function to determine the slope of the tangent line. The function is given by \( y = x \sqrt{2x + 3} \). Use the product rule for differentiation: if \( u = x \) and \( v = \sqrt{2x+3} \), then \( u' = 1 \) and \( v' = \frac{1}{2} (2x+3)^{-1/2} \cdot 2 \) (using the chain rule). Applying the product rule: \[ y' = u'v + uv' = 1 \cdot \sqrt{2x+3} + x \cdot \frac{1}{2}\cdot \frac{2}{\sqrt{2x+3}} \] This simplifies to: \[ y' = \sqrt{2x+3} + \frac{x}{\sqrt{2x+3}} \] Combine these terms under a common denominator (\( \sqrt{2x+3} \)): \[ y' = \frac{(2x+3) + x}{\sqrt{2x+3}} = \frac{3x + 3}{\sqrt{2x+3}} \] Thus, the simplified derivative is: \[ y' = \frac{3(x+1)}{\sqrt{2x+3}} \]

Step 2: Evaluate the Derivative at the Given Point

Now, let's find the slope of the tangent line by evaluating the derivative at \( x = 3 \). Substitute \( x = 3 \) into \( y' = \frac{3(x+1)}{\sqrt{2x+3}} \): \[ y'(3) = \frac{3(3+1)}{\sqrt{2\times3+3}} = \frac{3 \times 4}{\sqrt{6+3}} = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4 \] The slope of the tangent line at \( (3, 9) \) is 4.

Step 3: Find the Equation of the Tangent Line

With the slope from Step 2, use the point-slope form of the line equation: \[ y - y_1 = m(x - x_1) \] Here, \( m = 4 \) is the slope, and \((x_1, y_1) = (3, 9)\) is the point on the line. Substitute: \[ y - 9 = 4(x - 3) \] Distribute and simplify: \[ y - 9 = 4x - 12 \]Add 9 to both sides: \[ y = 4x - 3 \] Thus, the equation of the tangent line is \( y = 4x - 3 \).

Key Concepts

Tangent Line

Imagine you're driving along a gentle curve on a mountain road. At a specific point, the trajectory of your car aligns perfectly with a straight section of the road. This is much like a tangent line on a graph. A tangent line touches the curve of a function at just one point and has the same slope as the function at that specific location. The significance of the tangent line is that it gives a linear approximation of the function at a particular point. Essentially, it is the best straight-line approximation for the curve at that particular point. For instance, the process to find an equation for the tangent line involves two key components: evaluating the slope of the curve at that point, and then using that slope to find the equation of the tangent line. In the discussed exercise, we found the slope of the tangent line of the function at the point (3,9) by calculating the derivative. Subsequently, using the point-slope form of a line, we derived the equation of our tangent line, giving us a powerful tool for estimation and understanding the behavior of functions at specific points.

Product Rule

When we want to differentiate a function that is the product of two or more simpler functions, we employ the product rule. This rule states that if you have two functions multiplied together, say, a function \( u(x) \) and another \( v(x) \), then the derivative of their product is computed as: \[ (uv)' = u'v + uv' \].Using the product rule allows us to break down more complex derivatives into manageable parts. The important component here is recognizing which part of your equation is \( u \) and which is \( v \). In the given exercise, the function \( y = x \sqrt{2x + 3} \) needed to be broken down this way.By setting \( u = x \) and \( v = \sqrt{2x+3} \), and knowing the derivatives \( u' \) and \( v' \), we applied the product rule to find the first derivative of the function. This technique is foundational in understanding relationships between variables in calculus and simplifies what initially might appear complex functions.

Derivative

In calculus, a derivative is a measure of how a function changes as its input changes. It’s the essential concept for describing the rate of change and is foundational in areas such as physics and engineering. When you calculate the derivative of a function at a point, you are finding the slope of the tangent line at that point. This gives great insights into the behavior of functions, such as determining points of maximization or minimization and predicting future trends. In the exercise, taking the derivative was crucial in identifying the slope of the curve at the point (3,9). We then used this derivative to calculate the equation of the tangent line, showcasing how calculating derivatives is essential for solving real-world problems and optimizing scenarios.

Chain Rule

The chain rule is a powerful tool in calculus for finding derivatives of composite functions. When examining a function composed of multiple layers, the chain rule helps take the derivative correctly. Essentially, it tells you how to "chain" the effects of two or more functions together.If a function \( y \) is dependent on \( u \), which in turn is dependent on \( x \), the chain rule allows you to find \( \frac{dy}{dx} \) by first finding \( \frac{dy}{du} \) and \( \frac{du}{dx} \). In our problem, to differentiate \( v = \sqrt{2x+3} \) involved the chain rule. We had a function within a function, \( 2x+3 \) inside the square root. Applying the chain rule, we found \( v' \) by taking the derivative of the outer function and multiplying it by the derivative of the inner function. The chain rule is indispensable when tackling problems involving layers of functions.