Problem 9 - Calculus Single Variable

Question

The variable \(y\) is given as a function of \(x\), which depends on \(t\). The values \(y_{0}\) and \(s_{0}\) of, respectively, \(y\) and \(d y / d t\) are given at a value \(t_{0}\) of \(t\). Use this data to find \(d x / d t\) at \(t_{0}\). $$ y=6 \exp (-3 x), \quad y_{0}=6, \quad s_{0}=54 $$

Step-by-Step Solution

Answer

\( dx/dt = -3 \) at \( t = t_0 \).

Step 1: Differentiate y with respect to t

Since we know that \( y = 6 \exp(-3x) \) and \( y = y(x(t)) \), we can use the chain rule to differentiate with respect to \( t \). The chain rule gives: \[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \]. Differentiate \( y = 6 \exp(-3x) \) with respect to \( x \) to get: \[ \frac{dy}{dx} = -18 \exp(-3x) \]. So, \[ \frac{dy}{dt} = -18 \exp(-3x) \cdot \frac{dx}{dt} \].

Step 2: Evaluate derivatives at given values

We are given \( y_0 = 6 \) and \( dy/dt = s_0 = 54 \) at \( t = t_0 \). At \( y = y_0 = 6 \), substitute back into the function to find: \[ 6 = 6 \exp(-3x) \]. Solving gives \( \exp(-3x) = 1 \), hence \( x = 0 \) because the exponential function is 1 when the exponent is zero.

Step 3: Substitute known values into derivative equation

Substitute \( \exp(-3x) = 1 \), \( dy/dt = 54 \), and \( x = 0 \) into the differentiated function \[ \frac{dy}{dt} = -18 \exp(-3x) \cdot \frac{dx}{dt} \]. This gives: \[ 54 = -18 \cdot 1 \cdot \frac{dx}{dt} \].

Key Concepts

Chain Rule

When faced with composite functions, where one function acts upon another, the chain rule becomes our best friend. It's an essential technique in differentiation. Think of it like peeling an onion, layer by layer.
For a composite function like \( y = y(x(t)) \), the chain rule allows us to differentiate with respect to a third variable \( t \). By symbolically representing it, if you have \( y = f(g(x)) \), the chain rule tells us:

• First, differentiate the outer function \( f \) with respect to its argument \( g(x) \).
• Then, multiply it by the derivative of the inner function \( g(x) \) with respect to \( x \).

Nestled within our specific problem, \( y = 6 \exp(-3x) \) and \( x \) depends on \( t \). Therefore, when applying the chain rule here:

• We first differentiate \( y \) with respect to \( x \), yielding \( \frac{dy}{dx} = -18 \exp(-3x) \).
• Next, multiply this by \( \frac{dx}{dt} \), leading us to \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \).

Understanding and applying the chain rule correctly is crucial for such problems.

Exponential Function

The exponential function, especially \( \exp(x) \) or \( e^x \), is unique due to its consistent behavior after differentiation. When you differentiate an exponential function, it self-replicates, making calculus easier in some computations. If you have \( e^x \), differentiating it in terms of \( x \) remains \( e^x \).
In this problem, the function \( y = 6 \exp(-3x) \) includes an exponential part where a constant multiplier resides in the exponent, making differentiation slightly more challenging. When we differentiate \( \exp(-3x) \) with respect to \( x \), we apply the chain rule and find:

• The differentiation of the outer function \( 6\exp(-3x) \) requires taking out the multiplier \(-3\) due to the inner part, giving us \( \frac{dy}{dx} = -18\exp(-3x) \).

This power of exponentials is not only important in pure mathematical theory but finds utility in fields ranging from natural sciences to finance.

Implicit Differentiation

Implicit differentiation is a method used when it's difficult or impossible to separate variables on either side of an equation. It's a handy tool when direct differentiation isn't evident due to nested dependencies of the variables.
In our particular exercise, \( y \) is a function of \( x \), which itself depends on \( t \). This indirect dependency makes it challenging to solve using straightforward differentiation. However, implicit differentiation allows us to compute derivatives with respect to \( t \) by recognizing these relationships.
By applying implicit differentiation, you treat \( x \) as a hidden variable, acknowledging that its variation with respect to \( t \) affects \( y \). This provides:

• An expression relating \( \frac{dy}{dt} \), \( \frac{dy}{dx} \), and \( \frac{dx}{dt} \), which is: \( \frac{dy}{dt} = -18\exp(-3x)\frac{dx}{dt} \).

Implicit differentiation allows you to handle complex relationships between multiple variables much more easily. It becomes crucial when combined with the chain rule in such scenarios.

Calculus Problem Solving

Tackling a calculus problem involves understanding the problem's constraints and relationships between variables. It's like unraveling a mystery one clue at a time.
First, recognize the function and its dependencies. In this case, \( y = 6 \exp(-3x) \), with \( x \) depending on \( t \).

• Start by differentiating with respect to \( t \), using the chain rule and managing the exponential function.
• Solve for any variables using given values to make further progress. Given \( y_0 = 6 \) indicates \( \exp(-3x) = 1 \) leading to \( x = 0 \).
• Plug these details back into your derived derivative equations to solve for the unknown derivative, such as \( \frac{dx}{dt} \).

This structured approach transforms a complex narrative into systematic steps, achieving a solid understanding and solution. Employing techniques like implicit differentiation within such steps greatly aids the process.