Problem 10 - University Calculus: Early Transcendentals

Question

If \(r+s^{2}+v^{3}=12, d r / d t=4,\) and \(d s / d t=-3,\) find \(d v / d t\) when \(r=3\) and \(s=1\).

Step-by-Step Solution

Answer

\(\frac{dv}{dt} = \frac{1}{6}\)

Step 1: Differentiate the Equation

Given the equation \(r + s^2 + v^3 = 12\), we first differentiate all terms with respect to \(t\).- The derivative of \(r\) with respect to \(t\) is \(\frac{dr}{dt}\).- The derivative of \(s^2\) with respect to \(t\) is \(2s \frac{ds}{dt}\).- The derivative of \(v^3\) with respect to \(t\) is \(3v^2 \frac{dv}{dt}\).Thus, the differentiated equation is:\[\frac{dr}{dt} + 2s \frac{ds}{dt} + 3v^2 \frac{dv}{dt} = 0\]

Step 2: Substitute Known Values

We are given that \(r = 3\), \(s = 1\), \(\frac{dr}{dt} = 4\), and \(\frac{ds}{dt} = -3\). Substitute these values into the differentiated equation.\[4 + 2(1)(-3) + 3v^2 \frac{dv}{dt} = 0\]Simplify the equation:\[4 - 6 + 3v^2 \frac{dv}{dt} = 0\]This reduces to:\[-2 + 3v^2 \frac{dv}{dt} = 0\]

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

From \(-2 + 3v^2 \frac{dv}{dt} = 0\), solve for \(\frac{dv}{dt}\):Add 2 to both sides:\[ 3v^2 \frac{dv}{dt} = 2 \]Divide both sides by \(3v^2\):\[ \frac{dv}{dt} = \frac{2}{3v^2} \]

Step 4: Find the Value of \(v\) when \(r = 3\) and \(s = 1\)

From the original equation \(r + s^2 + v^3 = 12\), substitute \(r = 3\) and \(s = 1\):\[ 3 + 1^2 + v^3 = 12 \]Simplify the equation:\[ 3 + 1 + v^3 = 12 \]\[ v^3 = 8 \]Taking the cube root of both sides gives:\[ v = 2 \]

Step 5: Substitute \(v\) into the Derivative Equation

Now substitute \(v = 2\) into the equation for \(\frac{dv}{dt}\):\[ \frac{dv}{dt} = \frac{2}{3(2)^2} \]Calculate the expression:\[ \frac{dv}{dt} = \frac{2}{12} = \frac{1}{6} \]

Key Concepts

Derivatives

In calculus, derivatives represent how a function changes as its input changes. Think of it as a way to measure the rate of change or the slope of the curve of the function. In our problem, we're working with a multivariable function involving three variables: \(r\), \(s\), and \(v\). Each variable contributes to changes in the whole system. Here's a quick look at how derivatives come into play here:

• When differentiating \(r\) with respect to \(t\), \,\(\frac{dr}{dt}\) tells us how \(r\) changes over time.
• For \(s^2\), applying the chain rule gives us \,\(2s \frac{ds}{dt}\)\, indicating the rate of change managed by both \(s\) and its rate over time.
• Lastly, differentiating \(v^3\) as \,\(3v^2 \frac{dv}{dt}\)\, shows the dependency of \(v\)'s change rate on its own value \(v\).

Understanding these changes is crucial for solving related rates problems, allowing us to find the unknown derivatives when some rates are given.

Implicit Differentiation

Implicit differentiation is essential when you cannot easily solve an equation for one variable before differentiating. In the exercise, the equation is \(r + s^2 + v^3 = 12\), and each variable depends on time \(t\). We differentiated the whole equation with respect to \(t\) without solving for any variable explicitly:

• We applied the derivative to the entire equation and used the chain rule for each term involving a time-dependent variable.
• This means treating all variables as implicit functions of \(t\), even if they are not initially written that way.
• By doing this, we incorporated relationships between the rates at which each term changes and solved for the desired rate \(\frac{dv}{dt}\).

Implicit differentiation allows us to handle complex relationships in equations like the one we have, helping us find rates of change even when variables are heavily intertwined.

Differentiation with Respect to Time

Differentiation with respect to time deals with how variables change over time. It's particularly useful in physics and engineering where many processes develop over time. In our case:

• Variables \(r\), \(s\), and \(v\) change with time, making it necessary to find \(\frac{dv}{dt}\), which reveals how \(v\) changes as time progresses.
• Given \(\frac{dr}{dt} = 4\) and \(\frac{ds}{dt} = -3\), we know the change rates of \(r\) and \(s\). These help us find \(\frac{dv}{dt}\).
• This rate indicates how quickly \(v\) adjusts when \(r\) and \(s\) shift, crucial for understanding dynamic systems.

Differentiating with respect to time in related rates problems bridges changes in multiple variables, enabling you to compute how one aspect of a system reacts to instantaneous shifts in others.