Problem 15 - 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. $$ 2 x^{3}-x^{2} y+3 y=13, \quad P_{0}=(2,3), \quad v_{0}=-2 $$

Step-by-Step Solution

Answer

\(\frac{dy}{dt}\) at \(P_0\) is \(\frac{24}{7}\).

Step 1: Differentiate the Given Equation

Differentiate the equation with respect to time, \(t\):\[ \frac{d}{dt}(2x^3 - x^2y + 3y) = \frac{d}{dt}(13). \]Applying the product rule and chain rule, we get:\[ 6x^2 \frac{dx}{dt} - (2x \frac{dx}{dt} y + x^2 \frac{dy}{dt}) + 3 \frac{dy}{dt} = 0. \]

Step 2: Substitute Known Values

Substitute the given values into the differentiated equation. We have \(x = 2\), \(y = 3\), and \(v_0 = \frac{dx}{dt} = -2\):\[ 6(2)^2(-2) - (2(2)(-2)(3) + (2)^2 \frac{dy}{dt}) + 3 \frac{dy}{dt} = 0. \]

Step 3: Simplify the Equation

Simplify the equation:\[ 6(4)(-2) - (-8 \cdot 3 + 4 \frac{dy}{dt}) + 3 \frac{dy}{dt} = 0. \]This simplifies to:\[ -48 + 24 + 4 \frac{dy}{dt} + 3 \frac{dy}{dt} = 0. \]

Step 4: Solve for \(\frac{dy}{dt}\)

Combine like terms and solve for \(\frac{dy}{dt}\):\[ -24 + 7\frac{dy}{dt} = 0. \]\[ 7\frac{dy}{dt} = 24. \]\[ \frac{dy}{dt} = \frac{24}{7}. \]

Key Concepts

Rate of Change

In calculus, one of the main objectives is to understand how things change. When we talk about the "rate of change" for a function, we're really talking about how quickly or slowly the function is changing at any given point or over a certain interval.

In this exercise, the rate of change of the variables \(x\) and \(y\) with respect to time \(t\) is important. The rate of change of \(x\) is noted as \(\frac{dx}{dt}\) and is provided as \(-2\) at the point \(P_0 = (2, 3)\). This means that as time progresses, the value of \(x\) is decreasing. Our goal is to find the rate at which \(y\) is changing, denoted as \(\frac{dy}{dt}\).

Determine rates of change often involves differentiating equations implicitly, especially when multiple variables, like \(x\) and \(y\), are intertwined.

Product Rule

The product rule is a vital tool when it comes to differentiating equations that include multiple terms being multiplied together. For functions \(u(t)\) and \(v(t)\), the product rule states: \[\frac{d}{dt}[u(t)v(t)] = u(t)\frac{dv}{dt} + v(t)\frac{du}{dt}.\] In the context of our exercise, we have the term \(-x^2y\) that needs to be differentiated with respect to \(t\). Because \(x\) and \(y\) both depend on \(t\), the product rule is applied here.
Sometimes, identifying which part of the function is \(u(t)\) and which is \(v(t)\) can help administer the rule correctly, especially in more complex expressions."

By using this rule, the differentiated form of \(-x^2y\) becomes \[-(2x \frac{dx}{dt} y + x^2 \frac{dy}{dt}).\]This allows us to capture the interaction between the changes in \(x\) and \(y\) over time.

Chain Rule

The chain rule is another indispensable differentiation technique often used alongside the product rule in implicit differentiation problems. This rule is utilized when dealing with composite functions. If you have a function \(h(t) = g(f(t))\), then the chain rule states:\[\frac{dh}{dt} = g'(f(t)) \cdot f'(t),\]meaning we differentiate the outer function and multiply it by the derivative of the inner function.

In our scenario, the chain rule aids us especially in terms such as \(x^3\) and \(y\), where both \(x\) and \(y\) change as \(t\) changes:

• For \(2x^3\), applying the chain rule gives \(6x^2 \frac{dx}{dt}\).
• For \(3y\), it simply gives \(3\frac{dy}{dt}\) because \(y\) is a direct function of \(t\).

This application ensures each component's rate of change is appropriately included in the overall differentiation.