Problem 61 - Calculus for Biology and Medicine

Question

Assume that \(x\) and \(y\) are differentiable functions of \(t\). Find \(\frac{d y}{d t}\) when \(x^{2}+y^{2}=1, \frac{d x}{d t}=2\) for \(x=\frac{1}{2}\), and \(y>0 .\)

Step-by-Step Solution

Answer

\(\frac{dy}{dt} = -\frac{2\sqrt{3}}{3}\) when \(x = \frac{1}{2}\).

Step 1: Understand the Problem

We are given that \(x^2 + y^2 = 1\), which represents a circle of radius 1 centered at the origin. We need to find \(\frac{dy}{dt}\) using implicit differentiation, given \(\frac{dx}{dt} = 2\) at \(x = \frac{1}{2}\) with \(y > 0\).

Step 2: Differentiate Implicitly

Start by taking the derivative of both sides of the equation \(x^2 + y^2 = 1\) with respect to \(t\). Using the chain rule, the derivative will be: \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\).

Step 3: Solve for \(\frac{dy}{dt}\)

Rearrange the equation \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\) to solve for \(\frac{dy}{dt}\): \[ 2y\frac{dy}{dt} = -2x\frac{dx}{dt} \]Divide both sides by \(2y\) to isolate \(\frac{dy}{dt}\): \[ \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} \]

Step 4: Substitute Known Values

We know \(x = \frac{1}{2}\), \(\frac{dx}{dt} = 2\), and that \(y > 0\) satisfies \(x^2 + y^2 = 1\). First, find \(y\) when \(x = \frac{1}{2}\): \[ \left(\frac{1}{2}\right)^2 + y^2 = 1 \]\[ \frac{1}{4} + y^2 = 1 \]\[ y^2 = \frac{3}{4} \]\[ y = \frac{\sqrt{3}}{2} \] (since \(y > 0\)).

Step 5: Calculate \(\frac{dy}{dt}\)

Substitute \(x = \frac{1}{2}\), \(y = \frac{\sqrt{3}}{2}\), and \(\frac{dx}{dt} = 2\) into the formula for \(\frac{dy}{dt}\): \[ \frac{dy}{dt} = -\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} \times 2 \]Simplify the expression to find \(\frac{dy}{dt}\): \[ \frac{dy}{dt} = -\frac{1 \times 2}{\sqrt{3}} \]\[ \frac{dy}{dt} = -\frac{2}{\sqrt{3}} \]Rationalize the denominator: \[ \frac{dy}{dt} = -\frac{2\sqrt{3}}{3} \]

Step 6: Conclusion

The derivative \(\frac{dy}{dt}\) when \(x = \frac{1}{2}\) and \(y > 0\) is \(-\frac{2\sqrt{3}}{3}\).

Key Concepts

Differentiable Functions

When we say that functions are differentiable with respect to a variable, we mean that their derivatives can be calculated at every point within their domain. In this exercise, both \(x\) and \(y\) are stated to be differentiable functions of \(t\), which simply implies that the rate of change of \(x\) and \(y\) with respect to \(t\) can be expressed by derivatives: \(\frac{d x}{d t}\) and \(\frac{d y}{d t}\) respectively.

A differentiable function is smooth and continuous, with no breaks or sharp angles. This property is crucial because it allows us to use calculus tools such as the chain rule and implicit differentiation effectively.

So, when tackling problems involving such functions, remember that you can assume the existence of derivatives, and thus apply calculus techniques to find those derivatives. This forms the foundation for solving more complex real-world problems, such as calculating velocities and accelerations in physics or solving differential equations in engineering.

Chain Rule

The chain rule is a vital technique in calculus used for differentiating composite functions. The essence of the chain rule lies in taking the derivative of the outer function and multiplying it by the derivative of the inner function.

In this specific exercise, the chain rule helps us perform implicit differentiation on the equation \(x^2 + y^2 = 1\). Here's how it comes into play:

• Differentiating \(x^2\) with respect to \(t\) involves applying the chain rule to get \(2x \frac{dx}{dt}\).
• Similarly, for \(y^2\), it becomes \(2y \frac{dy}{dt}\).

Putting it all together, the equation becomes \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0\). In essence, implict differentiation coupled with the chain rule allows us to relate the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) in our problem. This technique is particularly useful in problems where variables are implicitly related, like those involving circles, ellipses, and other curves.

Rationalize the Denominator

Rationalizing the denominator is a standard practice in mathematics, used to simplify expressions that contain radicals in the denominator. This process makes the expressions clearer and easier to work with.

In the final step of the solution, the expression for \(\frac{dy}{dt}\) was \( -\frac{2}{\sqrt{3}} \). While useful, it is typically not ideal to have a square root in the denominator. To rationalize it, we multiply both the numerator and denominator by \(\sqrt{3}\):

• This changes the expression to: \(-\frac{2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \).
• Simplify it further to become \(-\frac{2\sqrt{3}}{3} \).

This technique makes the expression neater and often prepares it for further operations or interpretations in more advanced calculations. Always aim for a rationalized denominator in final answers to align with conventional mathematical standards.