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Chapter 2: Problem 13

Differentiate implicitly to find dy/dx. Then find the slope of the curve at the given point. $$x^{2} y-2 x^{3}-y^{3}+1=0 ; \quad(2,-3)$$

Short Answer

Expert verified
The slope at the given point (2, -3) is \( -\frac{36}{23} \)

Step by step solution

01

Differentiate the equation implicitly

Given the equation: \[ x^{2} y - 2 x^{3} - y^{3} + 1 = 0 \]Differentiate both sides of the equation with respect to \( x \). For each term, apply the product rule (\(\frac{d}{dx}(uv) = u'v + uv'\)) and the chain rule as needed: \[\frac{d}{dx}(x^{2} y) - \frac{d}{dx}(2 x^{3}) - \frac{d}{dx}(y^{3}) + \frac{d}{dx}(1) = 0\]This becomes:\[ 2x y + x^{2} \frac{dy}{dx} - 6x^2 - 3y^2 \frac{dy}{dx} = 0 \]
02

Isolate \( \frac{dy}{dx} \)

Rearrange the equation to group the terms involving \( \frac{dy}{dx} \):\[ x^{2} \frac{dy}{dx} - 3 y^{2} \frac{dy}{dx} = 6 x^2 - 2 x y \]Factor out \( \frac{dy}{dx} \):\[ \left( x^{2} - 3y^{2} \right) \frac{dy}{dx} = 6 x^{2} - 2 xy \]Solve for \( \frac{dy}{dx} \) by dividing both sides by \( x^{2} - 3 y^{2} \):\[ \frac{dy}{dx} = \frac{6 x^{2} - 2 x y}{x^{2} - 3 y^{2}} \]
03

Plug in the given point

Substitute the given point \((2, -3)\) into the derivative to find the slope at that point:\[ \frac{dy}{dx} \Bigg|_{(2, -3)} = \frac{6(2)^{2} - 2(2)(-3)}{(2)^{2} - 3(-3)^{2}} \]Simplify the numerator and the denominator:\[ \frac{dy}{dx} \Bigg|_{(2, -3)} = \frac{6(4) + 6(2)}{4 - 3(9)} = \frac{24 + 12}{4 - 27} = \frac{36}{-23} \]Thus, the slope at the given point \( (2, -3) \) is:\[ \frac{dy}{dx} = -\frac{36}{23} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
In calculus, the product rule is used when differentiating expressions that are the product of two functions. For example, if you have an expression like \(x^2 y\), where both \(x\) and \(y\) are variables, you need the product rule to differentiate it. The product rule formula is:
\[ \frac{d}{dx}(uv) = u'v + uv' \ \]
where \(u\) and \(v\) are functions of \(x\).
So, in the given exercise, for differentiating \(x^2 y\), we let:
  • \(u = x^2\)
  • \(v = y\)

Differentiating each function gives: \(u' = 2x\) and \(v' = \frac{dy}{dx}\). Applying the product rule, we get: \( \frac{d}{dx}(x^2 y) = 2x y + x^2 \frac{dy}{dx} \).
This becomes part of the implicit differentiation process.
Chain Rule
The chain rule helps in differentiating composite functions. When a function is nested within another function, you must use the chain rule.
The formula for the chain rule is:
\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x) \]

In our exercise, we apply the chain rule to differentiate functions like \(y^3\). Since \(y\) is a function of \(x\), differentiate \(y^3\) as:
  • First, differentiate the outer function: \(f(y) = y^3\) gives \(f'(y) = 3y^2\).
  • Then, multiply by the derivative of the inner function (\(y\)): \(\frac{dy}{dx}\)
Thus, \( \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx} \).
This step is crucial in implicit differentiation when dealing with composite functions.
Finding Slope
To find the slope of the curve at a particular point after differentiating implicitly, we substitute the point coordinates into the derivative expression.
In the given problem, the derivative found was:
\[ \frac{dy}{dx} = \frac{6 x^{2} - 2 x y}{x^{2} - 3 y^{2}} \]

At the point \((2, -3)\), plug in these values:
  • \(x = 2\)
  • \(y = -3\)

This results in:
\[ \frac{dy}{dx} \bigg|_{(2, -3)} = \frac{6(2)^{2} - 2(2)(-3)}{(2)^{2} - 3(-3)^{2}} \ \] Simplifying both the numerator and the denominator:
\[ \frac{dy}{dx} \bigg|_{(2, -3)} = \frac{24 + 12}{4 - 27} = \frac{36}{-23} \ \]
The slope at the point \((2, -3)\) is:
\( \frac{dy}{dx} = -\frac{36}{23} \).
Calculus Problems
Calculus problems like the one given test your understanding of concepts like implicit differentiation, the product rule, and the chain rule. It's important to:
  • Identify which rules apply to each term.
  • Implicitly differentiate complex equations by treating certain variables as functions of each other.
  • Carefully apply the product and chain rules where necessary.

Implicit differentiation is especially useful when the relationship between variables isn't easily solved for one variable in terms of another.
Once differentiation is complete, substituting given points helps in finding specific values like the slope at that point.
Mastering these concepts ensures you can tackle more advanced calculus problems, build intuition for mathematical relationships, and apply them to real-world situations.

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