Problem 14 - Calculus Single Variable

Question

Variables \(x\) and \(y,\) which depend on \(t,\) are related by a given equation. A point \(P_{0}\) on the graph of that equation is also given, as is one of the following two values: $$v_{0}=\left.\frac{d x}{d t}\right|_{P_{0}} \quad \text { or } \quad s_{0}=\left.\frac{d y}{d t}\right|_{P_{0}}$$ Find the other. $$ 2 \sqrt{x}-\sqrt{3 y}=1, \quad P_{0}=(4,3), \quad s_{0}=5 $$

Step-by-Step Solution

Answer

\( v_{0} = 5 \). Both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are equal at \( P_{0} \).

Step 1: Differentiate the Given Equation

Start by differentiating the given equation implicitly with respect to \( t \). The given equation is \( 2 \sqrt{x} - \sqrt{3y} = 1 \). Applying chain rule, differentiate each term: \( \frac{d}{dt}[2 \sqrt{x}] = 2 \cdot \frac{1}{2\sqrt{x}} \cdot \frac{dx}{dt} = \frac{1}{\sqrt{x}} \cdot \frac{dx}{dt} \) and \( \frac{d}{dt}[\sqrt{3y}] = \frac{1}{2\sqrt{3y}} \cdot 3 \cdot \frac{dy}{dt} = \frac{3}{2\sqrt{3y}} \cdot \frac{dy}{dt} \). Therefore, the differentiated equation is:

Step 2: Write the Differentiated Equation

The differentiated form of the equation \( 2 \sqrt{x} - \sqrt{3y} = 1 \) with respect to \( t \) is: \[ \frac{1}{\sqrt{x}} \cdot \frac{dx}{dt} - \frac{3}{2\sqrt{3y}} \cdot \frac{dy}{dt} = 0 \] This equation relates \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

Step 3: Substitute Known Values Into the Equation

Now, substitute the coordinates of the point \( P_{0} = (4, 3) \) and the known value \( s_{0} = 5 \) into the differentiated equation. \[ \frac{1}{\sqrt{4}} \cdot \frac{dx}{dt} - \frac{3}{2\sqrt{3 \cdot 3}} \cdot 5 = 0 \] This simplifies to: \[ \frac{1}{2} \cdot \frac{dx}{dt} - \frac{3}{6} \cdot 5 = 0 \]

Step 4: Solve for the Unknown Variable \( \frac{dx}{dt} \)

The simplified equation is: \[ \frac{1}{2} \cdot \frac{dx}{dt} - \frac{3}{6} \cdot 5 = 0 \] Simplifying further: \[ \frac{1}{2} \cdot \frac{dx}{dt} - \frac{5}{2} = 0 \] Solve for \( \frac{dx}{dt} \) by rearranging: \[ \frac{1}{2} \cdot \frac{dx}{dt} = \frac{5}{2} \] Multiply both sides by 2: \[ \frac{dx}{dt} = 5 \]

Step 5: Conclusion: Find the Unknown Value

The value of \( \frac{dx}{dt} \) at the point \( P_{0} \) is \( 5 \). Thus, the other value, \( v_{0} \), is \( 5 \).

Key Concepts

Related Rates

Related rates problems involve understanding how time-dependent variables change in relation to each other. In this exercise, we explore how variables "x" and "y" change over time according to a specified equation. These variables are not independent; instead, each is affected by the other through their relationship to time, marked as "t".
This study of related rates is powerful in real-world applications where multiple phenomena depend on a common variable. For instance, in physics, related rates might model the relationship between distance and velocity over time.
To solve these problems:

• Identify all variables and their dependencies.
• Differently, the equation using implicit differentiation.
• Plug known values to find unknown rates of change.

The key is to understand how variations in one variable's rate require adjustments to others to maintain the given equation's integrity.

Chain Rule

The chain rule is an essential differentiation technique applied when dealing with composite functions. It allows us to unravel the differentiation of functions within functions. In the exercise, variables "x" and "y" are expressed as functions of "t" - time.
Using the chain rule involves two main steps:

• Apply the chain rule to each function by multiplying the derivative of the outer function by the derivative of the inner function.
• In the case of our problem, differentiate the square roots and their respective coefficients as they depend on "t".

For instance, to differentiate \(2 \sqrt{x}\) and \(\sqrt{3y}\) with respect to "t", we apply the chain rule, resulting in terms involving \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) respectively. This helps capture each component's time-dependent rate of change and how they interrelate within the broader equation.

Differentiation Techniques

Differentiation techniques are the mathematical tools that allow us to find the rate at which one variable changes with respect to another. In the context of this exercise, we're interested in implicit differentiation, finding the relationship between "x" and "y"'s rates with respect to time "t".
Implicit differentiation is useful when you cannot easily solve for one variable in terms of another. For example, with the equation \(2 \sqrt{x} - \sqrt{3y} = 1\), direct algebraic manipulation to find "x" or "y" may be complex.
These techniques often involve:

• Differentiating both sides of an equation with respect to a third variable.
• Using rules like the product and chain rules where necessary.

Through implicit differentiation, the relationship between derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) in our problem can be utilized to solve for one unknown, given the other. It simplifies otherwise complex relationships into workable results.

Variable Dependency

Variable dependency occurs when variables rely on one another within an equation, rather than functioning independently. This concept is central to related rates problems, where the change over time of one variable affects others.
In our problem, both "x" and "y" depend on "t," and are connected through the equation \(2 \sqrt{x} - \sqrt{3y} = 1\). This implies any change in "x" will result in a necessary change in "y" to maintain the equation's equality.
To handle variable dependency:

• Acknowledge each variable's direct dependence on time or each other.
• Use differentiated equations to interpret how changes in one variable influence the other.

When solving for unknown values, recognizing these dependencies ensures accurate calculation and deeper insight into how connected systems operate.