Problem 9 - Calculus for Biology and Medicine

Question

Find \(\frac{d y}{d x}\) if \(\left(x^{2}+y^{2}\right) e^{y}=0\)

Step-by-Step Solution

Answer

The derivative \(\frac{dy}{dx}\) is undetermined at \((x, y) = (0, 0)\).

Step 1: Identify the Equation

The given equation is \((x^{2} + y^{2})e^{y} = 0\). This is an implicit equation involving both \(x\) and \(y\).

Step 2: Recognize the Zero Product Condition

Since \((x^{2} + y^{2})e^{y} = 0\), either \(x^{2} + y^{2} = 0\) or \(e^{y} = 0\). The exponential function \(e^{y}\) is never zero, so we focus on \(x^{2} + y^{2} = 0\).

Step 3: Solve the Zero Condition

The equation \(x^{2} + y^{2} = 0\) implies that both \(x = 0\) and \(y = 0\) because squares of real numbers are non-negative and their sum can only be zero if both are zero.

Step 4: Differentiate Implicitly

Differentiate both sides of \((x^{2} + y^{2})e^{y} = 0\) with respect to \(x\). Use the product and chain rules: \(\frac{d}{dx}[(x^2+y^2)e^y] = 0\).

Step 5: Apply the Product Rule

Using the product rule on \((x^2 + y^2)e^y\), we get \([(2x + 2yy')e^y + (x^2 + y^2)e^y y'] = 0\), where \(y' = \frac{dy}{dx}\).

Step 6: Simplify the Differentiated Equation

Since \((x^2 + y^2) = 0\) as found earlier, the equation simplifies to \([2x + 2yy'] = 0\).

Step 7: Solve for \(\frac{dy}{dx}\)

Rearrange the equation \([2x + 2yy'] = 0\) to solve for \(y'\): \(2yy' = -2x\)Divide by \(2y\), assuming \(yeq0\): \(y' = -\frac{x}{y}\). However, at the point where \(x = 0\) and \(y = 0\), this solution doesn't provide a specific slope and \(\frac{dy}{dx}\) is undetermined.

Key Concepts

Zero Product Property

The zero product property is a fundamental algebraic principle. It states that if the product of two or more factors is zero, then at least one of the factors must be zero. In the equation \((x^2 + y^2)e^y = 0\), we must find the conditions where the product equals zero.
Because the exponential function \(e^y\) is never zero for any real number \(y\), the only way for the product \((x^2 + y^2)e^y\) to equal zero is if \(x^2 + y^2 = 0\).
In the real number system, squares of numbers are always non-negative. Therefore, \(x^2 + y^2 = 0\) implies both \(x = 0\) and \(y = 0\). This understanding of the zero product property is essential for solving problems involving multiplication leading to zero.

Product Rule

The product rule is a crucial technique in calculus used for finding the derivative of the product of two functions. For two functions, \(u(x)\) and \(v(x)\), the product rule is: \[(uv)' = u'v + uv'\]In this case, we're dealing with the expression \((x^2 + y^2)e^y\).

• Let \(u = x^2 + y^2\) and \(v = e^y\).
• The derivative is \(u'v + uv'\).

For \(u = x^2 + y^2\), we differentiate each component: \(u' = 2x + 2yy'\), applying implicit differentiation.
For \(v = e^y\), the derivative \(v'\) is \(e^y y'\), following the chain rule.
Combining these, the product rule assists in decomposing the problem as \((2x + 2yy')e^y + (x^2 + y^2)e^y y' = 0\).
The proper use of the product rule is vital for solving complex differentiation problems involving products.

Chain Rule

The chain rule is an essential rule in calculus for differentiating composite functions. A composite function is one where a function is applied within another function, represented as \(f(g(x))\). The chain rule states:
If \(y = f(g(x))\), then \[\frac{dy}{dx} = f'(g(x)) \, g'(x)\] In the context of our problem, consider the term \(v = e^y\):

• The function can be looked at as \(f = e^y\) and a nested or inner part \(y = g(x)\).

When differentiating \(e^y\) while considering \(y\) as a function of \(x\), the chain rule provides:\( ve^y y'\).
This method ensures that we correctly evaluate derivatives for functions within functions and is especially useful for handling implicit differentiation scenarios, where both \(x\) and \(y\) are in one equation.

Undetermined Form

An undetermined form in calculus and algebra often arises when an expression leads to an undefined state or ambiguity in computations. In the problem, we end up with \(y'\) defined as \(-\frac{x}{y}\), which becomes undetermined under certain circumstances.
Specifically, when both \(x = 0\) and \(y = 0\), the fraction \(-\frac{x}{y}\) becomes \(-\frac{0}{0}\).
This is an undetermined form. Mathematicians often solve such indeterminate forms using careful analysis, considering limits or redefining conditions.
In many practical instances, additional context or information is necessary to evaluate these forms accurately, but in our exercise, they show that \(\frac{dy}{dx}\) cannot be uniquely determined at the point where both variables simultaneously equal zero.