Question

Find an equation of the tangent line to the function at the given point. Then find the function values and the tangent line values at \(f(x+\Delta x)\) and \(y(x+\Delta x)\) for \(\Delta x=-0.01\) and \(0.01\). \(f(x)=\sqrt{25-x^{2}}\) \((3,4)\)

Short answer

The equation of the tangent line to the function at point (3,4) is \(y = -3/2x + 13/2\). The pairs of function and tangent line values for \(\Delta x = -0.01\) and \(0.01\) are approximately \(f(2.99) \approx 4.005\), \(y(2.99) \approx 4.005\), \(f(3.01) \approx 3.995\), and \(y(3.01) \approx 3.995\), respectively.

Step-by-step solution

1. Find the derivative of the function

We need to find \(f'(x)\) using the given function \(f(x)=\sqrt{25-x^{2}}\). You can rewrite the function as \(f(x)=(25-x^{2})^{1/2}\) and use the chain rule to differentiate. This would give \(f'(x) = -x(25-x^{2})^{-1/2}\).

2. Find the slope of the tangent line at the given point

Substitute the x-coordinate of the given point (which is 3) to the derivative to get the slope of the tangent. By substituting \(x=3\) into \(f'(x) = -x(25-x^{2})^{-1/2}\), you get a slope of \(-3/2\).

3. Find the equation of the tangent line

Using the slope-point form of a line equation \(y - y_{1} = m(x - x_{1})\), where \(m\) is the slope and \((x_{1}, y_{1})\) is the given point, substitute \(m=-3/2\) and \((x_{1}, y_{1})=(3,4)\) to get the equation \(y - 4 = -3/2(x - 3)\), which simplifies to \(y = -3/2x + 13/2\).

4. Find the function and tangent line values for \(\Delta x = -0.01\) and \(0.01\)

Use \(\Delta x = -0.01\) and \(0.01\) at point \(x=3\) to find the function and tangent line values. Substituting the values to the function \(f(x)=\sqrt{25-x^{2}}\) and the tangent line equation \(y = -3/2x + 13/2\), we get the following pairs of function and tangent line values: for \(\Delta x=-0.01\), \(f(2.99) \approx 4.005\) and \(y(2.99) \approx 4.005\), and for \(\Delta x=0.01\), \(f(3.01) \approx 3.995\) and \(y(3.01) \approx 3.995\).

Key concepts

Derivative Calculation

Calculating a derivative essentially involves finding the rate at which a function changes at any given point. This rate of change is crucial for determining the behavior of mathematical functions.
To find the derivative of a function like \(f(x)=\sqrt{25-x^{2}}\), we rewrite it in a form that is easier to handle: \(f(x)=(25-x^{2})^{1/2}\). Using the chain rule, a fundamental rule in differentiation, we find that the derivative \(f'(x)\) is \(-x(25-x^{2})^{-1/2}\).
The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. This concept is vital because it allows us to handle more complex functions by breaking them down into simpler parts.
Recognizing the appropriate method and being able to apply these rules is a key skill in calculus.

Slope of Tangent Line

The slope of a tangent line to a curve at a specific point indicates how steep the line is. This concept is important because the slope provides a snapshot of the curve’s behavior at that exact point.
Once we have found the derivative \(f'(x)\), we can determine the slope of the tangent line at any point on the curve. In this case, we substitute the x-coordinate (3) from the given point \((3,4)\) into the derivative: \(-3(25-3^{2})^{-1/2}\).
As a result, the slope at this point is \(-3/2\). This negative slope indicates that the tangent line is declining as we move from left to right.
The understanding of how derivative calculations translate into the slope of the line provides insight into the instantaneous rate of change of the function's value.

Function Differentiation

Function differentiation is the process of finding the derivative of a function. This process reveals how the function’s outputs change in relation to its inputs.
The differentiation involves rules and techniques such as the chain rule, product rule, and quotient rule. These rules help break down complex functions into manageable components.
In the example of \(f(x)=\sqrt{25-x^{2}}\), differentiation with respect to \(x\) tells us about the change in \(y\) as \(x\) changes, which is represented by \(-x(25-x^{2})^{-1/2}\).
These changes are essential for understanding the nature of the function, especially when considering how much and in what direction changes occur, like predicting how a small change in \(x\), such as \(\Delta x\), affects the function.

Point-Slope Form of a Line

The point-slope form is a way to express the equation of a line when you know a point on the line and the line’s slope. This form is useful because it quickly helps in writing the equation of a tangent line.
The general point-slope equation is \(y - y_{1} = m(x - x_{1})\), where \(m\) is the slope and \((x_{1}, y_{1})\) are the point's coordinates on the line. Using it on our example, with \(m = -3/2\) and \((x_{1}, y_{1}) = (3,4)\), we get:
\[ y - 4 = -\frac{3}{2}(x - 3)\]
Simplifying, this becomes \(y = -3/2x + 13/2\).
This form is an efficient way to create line equations, as it directly plugs in the vital information needed, simplifying the process of finding how a tangent line behaves at a particular point.