Problem 48 - Calculus for Biology and Medicine

Question

In Problems , find the tangent line, in standard form, to \(y=\) \(f(x)\) at the indicated point. $$ y=3 \pi x^{5}-\frac{\pi}{2} x^{3}, \text { at } x=-1 $$

Step-by-Step Solution

Answer

The tangent line is \(-27\pi x + 2y = 11\pi\).

Step 1: Find the Derivative

To find the equation of the tangent line, we first need the derivative of the function. The derivative of the function \( y = 3 \pi x^5 - \frac{\pi}{2} x^3 \) is calculated using the power rule.The derivative, \( y' \), will be:\[ y' = \frac{d}{dx}(3\pi x^5 - \frac{\pi}{2} x^3) = 15\pi x^4 - \frac{3\pi}{2} x^2 \]

Step 2: Evaluate the Derivative at the Given Point

Substitute \( x = -1 \) into the derivative to find the slope of the tangent line at this point.\[ y'(-1) = 15\pi (-1)^4 - \frac{3\pi}{2} (-1)^2 = 15\pi - \frac{3\pi}{2} = \frac{27\pi}{2} \]

Step 3: Calculate the Function Value at the Given Point

Substitute \( x = -1 \) into the original function to find the y-coordinate of the point where the tangent touches the curve.\[ y(-1) = 3\pi (-1)^5 - \frac{\pi}{2} (-1)^3 = -3\pi + \frac{\pi}{2} = -\frac{5\pi}{2} \]

Step 4: Write the Equation of the Tangent Line

Using the point-slope form, \( y - y_1 = m(x - x_1) \), where \( m = \frac{27\pi}{2} \), \( x_1 = -1 \), and \( y_1 = -\frac{5\pi}{2} \):\[ y + \frac{5\pi}{2} = \frac{27\pi}{2}(x + 1) \]Simplify and write in the standard form, \( Ax + By = C \):\[ y + \frac{5\pi}{2} = \frac{27\pi}{2} x + \frac{27\pi}{2} \]Subtract \( \frac{27\pi}{2} x \) from both sides:\[ -\frac{27\pi}{2} x + y = \frac{11\pi}{2} \]Multiply the entire equation by 2 to eliminate fractions:\[ -27\pi x + 2y = 11\pi \]

Key Concepts

Derivative Calculation

Calculating the derivative is a crucial step to finding the tangent line since it provides the slope at any given point of the function. To find the derivative, we break down the function into parts and apply differentiation rules. Here, the function is given by:

• \( y = 3 \pi x^5 - \frac{\pi}{2} x^3 \)

The power rule is the most straightforward method used here. It states that for a term like \( x^n \), the derivative is \( nx^{n-1} \). Let's apply this:

• The derivative of \( 3 \pi x^5 \) becomes \( 15 \pi x^4 \) because \( 5 \times 3\pi = 15\pi \), and we decrease the power by one.
• The derivative of \(-\frac{\pi}{2} x^3\) is \(-\frac{3 \pi}{2} x^2\), applying the power rule similarly.

Combining these results, the derivative is:

• \[ y' = 15\pi x^4 - \frac{3\pi}{2} x^2 \]

Tangent Line Equation

Once we have the derivative, we are well on our way to finding the tangent line. The equation of the tangent line is derived through the point-slope form, which is expressed as:

• \( y - y_1 = m(x - x_1) \)

Here, \( m \) is the slope we find by evaluating the derivative at the point of interest. In our case, substituting \( x = -1 \) yields a slope of:

• \[ y'(-1) = \frac{27\pi}{2} \]

For the point through which the tangent passes, we substitute \( x = -1 \) in the original function to get:

• \( y(-1) = -\frac{5\pi}{2} \)

By inserting these values into the point-slope form, we get the equation:

• \[ y + \frac{5\pi}{2} = \frac{27\pi}{2}(x + 1) \]\

This is then rearranged to the standard form, resulting in:

• \[ -27\pi x + 2y = 11\pi \]\

Power Rule Differentiation

The power rule is a foundational technique in calculus for finding the derivative of functions in polynomial form. It simplifies the process of differentiation and is crucial for understanding calculus operations. Here's how it works:

• For a term \( x^n \), the derivative is simply \( nx^{n-1} \).
• The constant factor multiplies the result as in the differentiation of \( c \times x^n \), yielding \( c \times nx^{n-1} \).

This rule allows us to quickly compute derivatives without complex calculations, making it especially handy in problems involving polynomials or where terms are easily written as powers. In our example, the power rule enabled the simplification of the terms \( 3\pi x^5 \) to \( 15\pi x^4 \) and \( -\frac{\pi}{2} x^3 \) to \( -\frac{3\pi}{2} x^2 \). Such simplicity showcases the elegance and power of this differentiation method, underpinning problems across calculus. Remember: mastering the power rule opens the gateway to understanding more advanced calculus derivatives easily.