Problem 47 - Calculus for Biology and Medicine

Question

In Problems , find the tangent line, in standard form, to \(y=\) \(f(x)\) at the indicated point. $$ y=\frac{1}{\sqrt{2}} x^{2}-\sqrt{2}, \text { at } x=4 $$

Step-by-Step Solution

Answer

The tangent line at \( x=4 \) is \( 4\sqrt{2}x - y - 9\sqrt{2} = 0 \).

Step 1: Differentiate the function

To find the tangent line, we first need the derivative of the given function which is the slope of the tangent line at any point. The function is \( y = \frac{1}{\sqrt{2}} x^{2}-\sqrt{2} \). The derivative, \( f'(x) \), is obtained by using basic differentiation rules: \( f'(x) = \frac{d}{dx} \left( \frac{1}{\sqrt{2}} x^{2} - \sqrt{2} \right) = \frac{2}{\sqrt{2}} x = \sqrt{2} x \).

Step 2: Evaluate the derivative at the given point

Next, we substitute \( x = 4 \) into the derivative to find the slope of the tangent line at the given point. So, \( f'(4) = \sqrt{2} \times 4 = 4\sqrt{2} \). The slope of the tangent line at \( x = 4 \) is \( 4\sqrt{2} \).

Step 3: Calculate the y-coordinate of the point

We need to find the y-coordinate of the point on the curve where \( x = 4 \). Substitute \( x = 4 \) into the original function to get \( y(4) = \frac{1}{\sqrt{2}}(4)^{2} - \sqrt{2} = \frac{16}{\sqrt{2}} - \sqrt{2} = 8\sqrt{2} - \sqrt{2} = 7\sqrt{2} \). The point on the curve is \( (4, 7\sqrt{2}) \).

Step 4: Write the equation of the tangent line in point-slope form

Use the point-slope form of a line equation, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope found in Step 2, and \( (x_1, y_1) = (4, 7\sqrt{2}) \). Substitute to find \( y - 7\sqrt{2} = 4\sqrt{2}(x - 4) \). Simplifying gives: \( y - 7\sqrt{2} = 4\sqrt{2}x - 16\sqrt{2} \).

Step 5: Convert the equation to standard form

For the standard form \( Ax + By + C = 0 \), rearrange the equation from Step 4: \( 4\sqrt{2}x - y - 9\sqrt{2} = 0 \). Thus, the equation of the tangent line in standard form is \( 4\sqrt{2}x - y - 9\sqrt{2} = 0 \).

Key Concepts

Differentiation

Differentiation is a calculus technique used to find the rate at which a function is changing at any given point. In simpler terms, it helps us determine how steep the curve of a function is at a certain point. To perform differentiation, we take the derivative of a function. For a function like \( y = \frac{1}{\sqrt{2}} x^{2} - \sqrt{2} \), the derivative is calculated using basic differentiation rules.

If you consider the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \), you can find that the derivative of \( \frac{1}{\sqrt{2}} x^{2} \) is \( \frac{2}{\sqrt{2}} x \), which simplifies to \( \sqrt{2} x \).
This derivative represents the slope of the tangent at any point \( x \) on the curve.

This process is fundamental for understanding the behavior of functions and is a key tool used in finding tangent lines.

Slope of a Curve

The slope of a curve at a particular point gives us a sense of how the curve behaves at that specific location. It tells us whether the curve is ascending, descending, or flat. For our function \( y = \frac{1}{\sqrt{2}} x^{2} - \sqrt{2} \), the slope at any point can be found by evaluating its derivative.

The calculated derivative, \( f'(x) = \sqrt{2} x \), gives us the slope of the curve for any value of \( x \). When we plug in \( x = 4 \), the derivative \( f'(4) = \sqrt{2} \times 4 = 4\sqrt{2} \) gives the exact slope of the curve at \( x = 4 \).

• The slope here, \( 4\sqrt{2} \), indicates how rapidly the curve is rising at this point.

This slope is also the slope of the tangent line that touches the curve only at this single point.

Point-Slope Form

The point-slope form is a convenient way to write the equation of a line when you know a point on the line and its slope. The form is:
\[ y - y_1 = m(x - x_1) \]
where \( m \) is the slope, and \((x_1, y_1)\) is a specific point on the line.

For this exercise, with a slope of \( 4\sqrt{2} \) and a point \((4, 7\sqrt{2})\), we substitute these values into the point-slope formula:

• \( y - 7\sqrt{2} = 4\sqrt{2}(x - 4) \)

This equation represents the tangent line in point-slope form, offering a clear way to see the line's characteristics by expressing how changes in \( x \) affect \( y \), starting from the known point \((4, 7\sqrt{2})\).
It's a straightforward representation when working with linear relationships and tangent lines.

Standard Form of a Line

The standard form of a line is another way to express a linear equation. This form helps to highlight the relationship between two variables \( x \) and \( y \) and can be a bit more formal than the point-slope form.

It's given as:

• \( Ax + By + C = 0 \)

where \( A \), \( B \), and \( C \) are constants.Using the last step of our calculation, we can rearrange the equation from point-slope form:

• \( y - 7\sqrt{2} = 4\sqrt{2}x - 16\sqrt{2} \)

to the standard form:

• \( 4\sqrt{2}x - y - 9\sqrt{2} = 0 \)

This equation encapsulates the tangent line's attributes in a clear, algebraic framework, making it easier to identify coefficients and manipulate them in broader mathematical contexts.