Question

If \(y^{3}-x^{2} y=8\) is the equation of a curve, find the slope and the equation of the tangent line at the point \((3,-1)\).

Short answer

The slope is 1 and the equation of the tangent line is \( y = x - 4 \).

Step-by-step solution

Differentiate implicitly

Given the equation of the curve: \[ y^3 - x^2 y = 8 \]Differentiate both sides of the equation with respect to x. Remember to use the product rule and chain rule where necessary.\[ \frac{d}{dx} (y^3) - \frac{d}{dx} (x^2 y) = \frac{d}{dx} (8) \]This results in:\[ 3y^2 \frac{dy}{dx} - (2x y + x^2 \frac{dy}{dx}) = 0 \]

Solve for \(\frac{dy}{dx}\)

Rearrange the differentiated equation to isolate \(\frac{dy}{dx}\):\[ 3y^2 \frac{dy}{dx} - 2xy - x^2 \frac{dy}{dx} = 0 \]Combine like terms involving \(\frac{dy}{dx}\):\[ (3y^2 - x^2) \frac{dy}{dx} = 2xy \]Finally, solve for \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = \frac{2xy}{3y^2 - x^2} \]

Plug in the point (3, -1)

Simplify \(\frac{dy}{dx}\) by plugging in the coordinates of the point (3, -1):\[ y = -1 \text{ and } x = 3 \]\[ \frac{dy}{dx} = \frac{2(3)(-1)}{3(-1)^2 - (3)^2} = \frac{-6}{3 - 9} = \frac{-6}{-6} = 1 \]Therefore, the slope of the tangent line at the point (3, -1) is 1.

Write the equation of the tangent line

Use the point-slope form of the line equation: \( y - y_1 = m(x - x_1) \)Given \( m = 1 \), \( x_1 = 3 \), and \( y_1 = -1 \), substitute these values into the equation:\[ y - (-1) = 1(x - 3) \]Simplify to get:\[ y + 1 = x - 3 \]\[ y = x - 4 \]Therefore, the equation of the tangent line at the point (3, -1) is \( y = x - 4 \).

Key concepts

Implicit Differentiation

Implicit differentiation is a technique used in calculus to find the derivative of a function when it is not explicitly solved for one variable in terms of another. This method becomes essential when dealing with equations where y is defined implicitly in terms of x—such as in our exercise.
In our example, the equation is given as: \[ y^3 - x^2 y = 8 \]
Rather than solving for y and then differentiating, we differentiate directly while treating y as a function of x. This requires using the chain rule and often the product rule. Here's the general approach:

• Differentiate both sides of the equation with respect to x, treating y as a function of x (i.e., differentiate y as you would with x but add dy/dx as a multiplier).
• Use the product rule where necessary, for example in products like x^2y.
• Solve for dy/dx by isolating terms and factoring when necessary.

This method allows us to find the relationship between the differentials of x and y even when y is not explicitly isolated.

Point-Slope Form

Once we've found the derivative (slope) at the given point, the next step is to write the equation of the tangent line at that point. The point-slope form of a line's equation is particularly useful for this purpose. The formula is: \[ y - y_1 = m(x - x_1) \]
Here,

• \ y_1 and x_1 \ are the coordinates of the given point (in our case, (3, -1)).
• m is the slope we've found using implicit differentiation (which is 1 for our exercise).

By plugging these values into the formula, we can directly write the equation of the tangent line passing through our given point. For our example: \[ y - (-1) = 1(x - 3) \]
Simplifying it results in the equation: \[ y = x - 4 \]
Thus, we easily find the tangent line at the specific point, making this form particularly practical for such exercises.

Calculus

In calculus, we study how functions change and how these changes can be described and analyzed using derivatives and integrals. The exercise prominent here revolves around derivatives, specifically finding the slope of a tangent line to a curve at a particular point.
Calculus concepts that are key to this process include:

• Derivatives: These measure the rate at which a function changes as its input changes. Implicit differentiation helps us find such rates even when the functions are not explicitly defined.
• Product Rule: Used when differentiating expressions where two variables are multiplied together, like \ x^2y \ in our equation. The product rule states that \[ \frac{d}{dx}(uv) = u'v + uv' \]
• Chain Rule: This comes into play when differentiating composite functions. For instance, differentiating \ y^3 \ involves using the chain rule because we treat y as a function of x.

By mastering these concepts, we can handle a range of problems involving changing rates and slopes, which are foundational to calculus.