Problem 12 - Precalculus Mathematics for Calculus

Question

Find an equation of the tangent line to the curve at the given point. Graph the curve and the tangent line. $$y=\frac{1}{x^{2}} \quad \text { at }(-1,1)$$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = 2x + 3 \).

Step 1: Find the Derivative of the Curve

To find the equation of the tangent line, we first need the derivative of the curve, which represents the slope of the tangent line at any given point. The function given is \( y = \frac{1}{x^2} \). We can rewrite this as \( y = x^{-2} \). Using the power rule, \( \frac{d}{dx}[x^n] = nx^{n-1} \), the derivative \( y' \) is \( -2x^{-3} \), or \( y' = -\frac{2}{x^3} \).

Step 2: Evaluate the Derivative at the Point

Next, we compute the derivative at the point \((-1, 1)\) to find the slope of the tangent line. Substituting \( x = -1 \) into the derivative \( y' = -\frac{2}{x^3} \), we get \( y'(-1) = -\frac{2}{(-1)^3} = 2 \). This means the tangent line has a slope of 2 at \( (-1, 1) \).

Step 3: Write the Equation of the Tangent Line

Having the slope \( m = 2 \) and using the point \((-1, 1)\), we can use the point-slope form of a line equation: \( y - y_1 = m(x - x_1) \). Substituting the given values, we find: \( y - 1 = 2(x + 1) \). Simplifying this, \( y = 2x + 2 + 1 \). Therefore, the equation of the tangent line is \( y = 2x + 3 \).

Step 4: Graph the Curve and the Tangent Line

To complete the exercise, plot the graph of \( y = \frac{1}{x^2} \) and the tangent line \( y = 2x + 3 \) on the same set of axes. The curve is a hyperbola that approaches the x-axis and y-axis asymptotically. The tangent line \( y = 2x + 3 \) should be straight and intersect the curve at the point \((-1, 1)\).

Key Concepts

Derivative

The concept of a derivative is the foundation of calculus, expressing how a function changes as its input changes. In simpler terms, it tells us the rate of change or the slope of a curve at a given point. For any function, the derivative provides a new function that describes these rates of change for every point along the curve.

When calculating the derivative, we often look for how a small change in the input (usually denoted as \(x\)) leads to a change in the output (denoted as \(y\)). This is especially useful for finding tangent lines, as these lines only touch the curve at a specific point and their slope is exactly the slope of the curve at that point.

• In our exercise, we found the derivative of \( y = \frac{1}{x^2} \), rewritten as \( y = x^{-2} \), by applying calculus techniques.
• The derivative tells us how the function \( y = \frac{1}{x^2} \) changes at any particular \(x\)-value.

Power Rule

One of the most essential tools in finding derivatives is the power rule. The power rule provides a quick way to find the derivative of functions of the form \(x^n\), where \(n\) is any real number. The rule states:\[ \frac{d}{dx}[x^n] = nx^{n-1} \]This means that if we have a power of \(x\), we can "bring down" the exponent as a coefficient and reduce the exponent by one.

In the given exercise, we had to find the derivative of \(y = \frac{1}{x^2}\) which can be expressed as \(y = x^{-2}\). By applying the power rule:

• We bring down the exponent \(-2\) as a coefficient.
• We subtract one from the exponent to get \(-2 - 1\), simplifying our expression to \(-2x^{-3}\).

Thus, we find that the derivative is \(y' = -\frac{2}{x^3}\). This formula gives us the slope of the tangent line at any point • \(x\)• on the curve.

Point-Slope Form

The point-slope form is a technique used in algebra to find the equation of a line given a point on the line and the slope of the line. The formula for the point-slope form is:\[ y - y_1 = m(x - x_1) \]where \((x_1, y_1)\) is a known point on the line and \(m\) is the slope. This form is particularly handy when dealing with tangent lines in calculus since we first find the slope using the derivative.

In our specific task, we calculated the slope at point \((-1, 1)\), which was 2. With this information:

• We plug in \((-1, 1)\) and the slope 2 into the point-slope form:
• \(y - 1 = 2(x + 1)\).

By simplifying, we rearrange it into the slope-intercept form \(y = mx + b\), and discover that the tangent line equation is \(y = 2x + 3\). The point-slope form efficiently builds a bridge from derivative calculation directly to the line equation, linking algebra with calculus seamlessly.