Problem 64

Question

Assume that $u$ and $v$ are differentiable functions of $t$. Find $\frac{d u}{d t}$ when $u^{2}+v^{3}=12, \frac{d v}{d t}=2$ for $v=2$, and $u>0$.

Step-by-Step Solution

Verified

Answer

$\frac{du}{dt} = -6$ when $v = 2$ and $u > 0$.

1Step 1: Understand the Problem

We are given the equation $u^2 + v^3 = 12$, and need to find $\frac{du}{dt}$ under the condition that $\frac{dv}{dt} = 2$ at $v = 2$, with $u > 0$.

2Step 2: Differentiate the Equation with Respect to t

Differentiate both sides of the equation $u^2 + v^3 = 12$ with respect to $t$. Using the chain rule, we get $2u \frac{du}{dt} + 3v^2 \frac{dv}{dt} = 0$.

3Step 3: Substitute Known Values

Substitute $v = 2$ and $\frac{dv}{dt} = 2$ into the differentiated equation to simplify. This gives: $2u \frac{du}{dt} + 3(2)^2(2) = 0$.

4Step 4: Simplify the Equation

Calculate the terms: $3(2)^2 = 12$ and $12 \cdot 2 = 24$. Substitute back to get $2u \frac{du}{dt} + 24 = 0$.

5Step 5: Solve for $\frac{du}{dt}$

Rearrange the equation $2u \frac{du}{dt} = -24$ and solve for $\frac{du}{dt}$ to obtain $\frac{du}{dt} = \frac{-24}{2u}$.

6Step 6: Find the Value of u at v=2

Go back to the original equation $u^2 + v^3 = 12$ and substitute $v = 2$ to find $u$: $u^2 + 8 = 12$ leads to $u^2 = 4$. Since $u > 0$, $u = 2$.

7Step 7: Final Calculation of $\frac{du}{dt}$

Substitute $u = 2$ into $\frac{du}{dt} = \frac{-24}{2u}$ to find $\frac{du}{dt} = \frac{-24}{4} = -6$.

Key Concepts

Chain RuleDifferentiable FunctionsRate of Change

Chain Rule

Implicit differentiation often uses a powerful tool called the **Chain Rule**. Let's see how it works in our problem. For the given equation $ u^2 + v^3 = 12 $, both $ u $ and $ v $ are functions of $ t $.
The Chain Rule helps us differentiate complex functions by breaking them down into more manageable parts. Essentially, it allows us to take the derivative of a composite function: a function within another function. In this scenario, since both $ u $ and $ v $ depend on $ t $, when differentiating their terms with respect to $ t $, we multiply by the derivative of these inner functions.
Specifically, for the term $ u^2 $, its derivative is calculated using the Chain Rule as follows:

Differentiate the outer function: $ 2u $.
Multiply by the derivative of the inner function: $ \frac{du}{dt} $.

This gives us $ 2u \frac{du}{dt} $. Similarly, for $ v^3 $:

Differentiate the outer function: $ 3v^2 $.
Multiply by the derivative of the inner function: $ \frac{dv}{dt} $.

This results in $ 3v^2 \frac{dv}{dt} $. Applying the Chain Rule effectively allows us to find rates of change within interconnected systems.

Differentiable Functions

Differentiability is a crucial property for functions involved in calculus. When we say a function is **differentiable**, it means that at every point in its domain, it has a well-defined derivative. In other words, the function's graph is smooth and not broken or kinked.
In our exercise, both $ u $ and $ v $ are specified to be differentiable functions of $ t $. This ensures that when we differentiate $ u^2 + v^3 = 12 $ with respect to $ t $, using implicit differentiation, we have valid derivatives $ \frac{du}{dt} $ and $ \frac{dv}{dt} $.
Smooth, continuous functionalities:

The existence of $ \frac{dv}{dt} = 2 $ highlights that $ v $ changes smoothly with respect to $ t $.
The process helps us to fathom $ \frac{du}{dt} $ as well, as there are no sharp turns or discontinuities impacting $ u $.

Being differentiable is essential for applying most calculus techniques, as discontinuities or abrupt changes would invalidate many derivative calculations.

Rate of Change

Understanding the **Rate of Change** is fundamental in interpreting derivatives.
Derivatives in essence tell us how one quantity changes with respect to another. In this context, $ \frac{du}{dt} $ and $ \frac{dv}{dt} $ represent the rates at which $ u $ and $ v $ change over time.
In the exercise, we needed to find $ \frac{du}{dt} $ given specific values to $ v $ and its rate of change. Since $ \frac{dv}{dt} = 2 $, it indicates $ v $ is increasing at a consistent pace. Our goal was to determine how this affects $ u $ — specifically, what $ \frac{du}{dt} $ is when $ v = 2 $.Steps to nail down the rate:

Using derivatives, we engage with the relationship between variables $ u $ and $ v $ as entwined by the equation $ u^2 + v^3 = 12 $.
After applying the Chain Rule and simplifying, we derived $ \frac{du}{dt} = \frac{-24}{2u} $.
Replacing $ u $ with its corresponding value of 2 results in a readily computable rate $ \frac{du}{dt} = -6 $, exemplifying a decrease in $ u $ as $ t $ progresses.

Understanding these rates helps us predict how variables interact and change over time, which is vital for dynamic systems modeling.

Problem 64

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