Problem 59 - Calculus and its applications

Question

Find the equation of the tangent line to the graph of \(g(x)=\sqrt[3]{x^{2}}\) a) at (-1,1)\(;\) b) at (1,1)\(;\) c) at (8,4) .

Step-by-Step Solution

Answer

a) \(y = \frac{2}{3}x + \frac{5}{3}\); b) \(y = \frac{2}{3}x + \frac{1}{3}\); c) \(y = \frac{1}{6}x + \frac{10}{3}\)."

Step 1: Find the derivative of the function

To find the equation of the tangent line, we first need the derivative. Given \(g(x) = \sqrt[3]{x^2} = (x^2)^{1/3}\), we use the chain rule and the power rule to differentiate: \(g'(x) = \frac{2}{3}x^{-1/3}(x^2)^{-1/3}\times 2x = \frac{2x}{3}\cdot\frac{1}{\sqrt[3]{x^4}} = \frac{2}{3\sqrt[3]{x^2}}\).

Step 2: Evaluate the derivative at the given points

Now we evaluate \(g'(x)\) at each point given. For \((-1, 1)\), \(g'(-1) = \frac{2}{3\sqrt[3]{1}} = \frac{2}{3} \).For \((1, 1)\), \(g'(1) = \frac{2}{3\sqrt[3]{1}} = \frac{2}{3} \).For \((8, 4)\), \(g'(8) = \frac{2}{3\sqrt[3]{64}} = \frac{2}{3 \times 4} = \frac{1}{6}\).

Step 3: Determine the equation of the tangent line

Use the point-slope form for the equation of a tangent line: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope from the derivative and \((x_1, y_1)\) is the point.For \((-1, 1)\), \(y - 1 = \frac{2}{3}(x + 1)\), simplifying to \(y = \frac{2}{3}x + \frac{5}{3}\).For \((1, 1)\), \(y - 1 = \frac{2}{3}(x - 1)\), simplifying to \(y = \frac{2}{3}x + \frac{1}{3}\).For \((8, 4)\), \(y - 4 = \frac{1}{6}(x - 8)\), simplifying to \(y = \frac{1}{6}x + \frac{10}{3}\).

Key Concepts

Derivative Calculation

Finding the derivative is a key step when determining the equation of a tangent line. For the function \( g(x) = \sqrt[3]{x^2} \), this becomes a bit intricate, but manageable with the right techniques.
Here, we use both the chain rule and the power rule. The rewritten function is \( (x^2)^{1/3} \).
This allows us to differentiate using these rules:

• Power Rule: Differentiate \( x^n \) as \( nx^{n-1} \)
• Chain Rule: Differentiate the outer function and multiply by the derivative of the inner function.

Combining these, we get the derivative:
\( g'(x) = \frac{2x}{3\sqrt[3]{x^4}} \), simplified to \( \frac{2}{3\sqrt[3]{x^2}} \).
This process may look complex at first, but breaking it down into smaller steps helps in understanding how each rule applies.

Equation of Tangent Line

Once we have the derivative, finding the tangent line's equation involves substituting into the point-slope form. This form, \( y - y_1 = m(x - x_1) \), connects the slope and point directly on the graph.
The slope \( m \) comes from the derivative value at the specific \( x \)-coordinate of the point. Evaluating the derivative at given points is straightforward:

• For \((-1, 1)\), \( g'(-1) = \frac{2}{3} \)
• For \((1, 1)\), \( g'(1) = \frac{2}{3} \)
• For \((8, 4)\), \( g'(8) = \frac{1}{6} \)

By inserting these slopes and the respective points into the equation, we derive tangent lines for each case. This demonstrates how each tangent fits uniquely onto the curve at the points specified.
With a clear view on inserting values into the point-slope form, the equation of tangent lines can be easily obtained.

Point-Slope Form

The point-slope form is a handy tool in calculus for finding lines quickly and accurately. It looks like this: \( y - y_1 = m(x - x_1) \).
This form helps when you have a point and a slope. The point \((x_1, y_1)\) represents position on the line, while \(m\) is the slope.
Using point-slope form in tangent line problems:

• With slope \( \frac{2}{3} \) at \((-1, 1)\): \( y - 1 = \frac{2}{3}(x + 1) \right \)
• With slope \( \frac{2}{3} \) at \((1, 1)\): \( y - 1 = \frac{2}{3}(x - 1) \right \)
• With slope \( \frac{1}{6} \) at \((8, 4)\): \( y - 4 = \frac{1}{6}(x - 8) \right \)

The simplifications lead to straight-line equations specific to each tangent.
Because of the intuitive layout of point-slope form, it becomes clear and direct for connecting points and slopes in problems of this nature.