Problem 47 - Calculus for Biology and Medicine

Question

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 in standard form is \( 4\sqrt{2}x - y = 9\sqrt{2} \).

Step 1: Find the derivative of the function

First, we need to differentiate the function to find the slope of the tangent line at the given point. The function is \( y = \frac{1}{\sqrt{2}}x^2 - \sqrt{2} \). Differentiating with respect to \( x \) gives us the derivative: \( f'(x) = 2 \times \frac{1}{\sqrt{2}}x = \frac{2}{\sqrt{2}}x = \sqrt{2}x \).

Step 2: Evaluate the derivative at the given point

Now, we need to find the slope of the tangent line at the point where \( x = 4 \). Substitute \( x = 4 \) into the derivative: \( f'(4) = \sqrt{2} \times 4 = 4\sqrt{2} \). This is the slope \( m \) of the tangent line.

Step 3: Find the y-coordinate of the point on the curve

To find the y-coordinate of the point, substitute \( x = 4 \) into the original function: \( y = \frac{1}{\sqrt{2}}(4)^2 - \sqrt{2} = \frac{1}{\sqrt{2}} \times 16 - \sqrt{2} = 8\sqrt{2} - \sqrt{2} = 7\sqrt{2} \). Thus, the point is \((4, 7\sqrt{2})\).

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

The point-slope form of a line is \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point on the line and \( m \) is the slope. Substituting \((4, 7\sqrt{2})\) and \( m = 4\sqrt{2} \), we get: \( y - 7\sqrt{2} = 4\sqrt{2}(x - 4) \).

Step 5: Convert the equation to standard form

The standard form of a line is \( Ax + By = C \). Expanding \( y - 7\sqrt{2} = 4\sqrt{2}x - 16\sqrt{2} \) gives \( y = 4\sqrt{2}x - 16\sqrt{2} + 7\sqrt{2} \). Therefore, \( y = 4\sqrt{2}x - 9\sqrt{2} \). Rewriting in standard form: \( -4\sqrt{2}x + y = -9\sqrt{2} \). Multiplying through by \( -1 \) gives the final equation: \( 4\sqrt{2}x - y = 9\sqrt{2} \).

Key Concepts

Derivative

Understanding the derivative is crucial when finding the tangent line to a curve. A derivative represents how a function changes as its input changes, essentially providing the slope of the function at any given point. In this problem, we have the function \( y = \frac{1}{\sqrt{2}}x^2 - \sqrt{2} \). To find the slope of the tangent line at \( x = 4 \), we need to differentiate this function with respect to \( x \).Differentiating the function, we use the power rule, \( \frac{d}{dx}x^n = nx^{n-1} \). Thus, the derivative is \( f'(x) = 2 \times \frac{1}{\sqrt{2}}x = \sqrt{2}x \). This derivative tells us how the function \( f(x) \) changes at any point \( x \).We then evaluate this derivative at the desired point by substituting \( x = 4 \):

• \( f'(4) = \sqrt{2} \times 4 = 4\sqrt{2} \)

This means, at \( x = 4 \), the slope of the tangent line is \( 4\sqrt{2} \).

Point-Slope Form

Once the slope of a tangent line is known, the next step is to express the tangent line in what is called the point-slope form. This form is particularly useful because it directly utilizes the slope and a specific point on the line. The point-slope form of a line is given by:

• \( y - y_1 = m(x - x_1) \)

where \( m \) is the slope and \( (x_1, y_1) \) is a point on the line. For our specific problem, the slope \( m \) is \( 4\sqrt{2} \), and the point \( (x_1, y_1) \) is \( (4, 7\sqrt{2}) \). When we substitute these into the point-slope formula, we get:

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

This equation gives us a neat representation of the tangent line using the known slope and point.

Standard Form

After obtaining the equation of the tangent line in point-slope form, the next step is to convert it into the standard form, which is a common requirement for representing line equations. Standard form is given by:

• \( Ax + By = C \)

This form can be easier to work with for certain applications and is often preferred for its conciseness.Starting from our point-slope equation:

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

We simplify and rearrange this equation:

• Expanding gives \( y = 4\sqrt{2}x - 16\sqrt{2} + 7\sqrt{2} \)
• Simplifying results in \( y = 4\sqrt{2}x - 9\sqrt{2} \)

Finally, we rearrange to standard form:

• \( -4\sqrt{2}x + y = -9\sqrt{2} \)
• Multiply by \(-1\) to get \( 4\sqrt{2}x - y = 9\sqrt{2} \)

Thus, the standard form of the tangent line is \( 4\sqrt{2}x - y = 9\sqrt{2} \).

Slope

Slope is a fundamental concept when working with lines, including tangent lines. In essence, slope measures the steepness or inclination of a line, describing how much \( y \) changes for a given change in \( x \). Mathematically, it is defined as "rise over run," the ratio of the vertical change to the horizontal change between two points on a line.In the context of tangent lines especially, the slope at a particular point on a curve reflects how the curve behaves locally at that point. For the curve \( y = \frac{1}{\sqrt{2}}x^2 - \sqrt{2} \), the slope of the tangent line at \( x = 4 \) was determined to be \( 4\sqrt{2} \). This tells us how quickly the y-value of the function is changing at this x-value and forms the basis for defining the tangent line:

• High slope values indicate a steeper line
• A slope of zero would indicate a flat line
• Negative slopes suggest the line descends from left to right

Understanding slope helps in visualizing how tangent lines interact with curves at specific points.