Problem 22 - Applied Calculus

Question

If \(U\) and \(V\) are positive constants, find all critical points of $$F(t)=U e^{t}+V e^{-t}$$

Step-by-Step Solution

Answer

The critical point is at \( t = \frac{1}{2} \ln\left(\frac{V}{U}\right) \).

Step 1: Find the derivative

To find the critical points, first we need to find the derivative of the function \[ F(t) = U e^t + V e^{-t}.\] The derivative using basic rules for derivatives of exponential functions is:\[ F'(t) = U e^t - V e^{-t}.\]

Step 2: Set the derivative equal to zero

Critical points occur where the derivative is zero. So, we set \[ F'(t) = 0: \U e^t - V e^{-t} = 0.\]

Step 3: Solve the equation for zero derivative

First, rearrange the equation:\[ U e^t = V e^{-t}.\]Multiply both sides by \( e^t \) to eliminate the negative exponent:\[ U e^{2t} = V.\]Then, solve for \(e^{2t}:\)\[ e^{2t} = \frac{V}{U}.\]Now log both sides\[ 2t = \ln\left(\frac{V}{U}\right).\]Finally, solve for \(t:\)\[ t = \frac{1}{2} \ln\left(\frac{V}{U}\right). \]

Step 4: Verify solution is a critical point

Replace the value of \(t\) back into the derivative,\[ F'(t) = U e^t - V e^{-t},\]to ensure it equals zero confirming it as a critical point.Plugging \(t = \frac{1}{2} \ln\left(\frac{V}{U}\right),\)the calculations will simplify to zero based on our prior derivations.

Key Concepts

Exponential Functions

Exponential functions are a fundamental type of function in calculus, characterized by a constant base raised to a variable exponent. A basic form is \( e^t \), where \( e \) is an irrational constant approximately equal to 2.71828. In our problem, we have both positive and negative exponents: \( Ue^t \) and \( Ve^{-t} \). These are classic examples of exponential functions, where \( U \) and \( V \) represent specific multipliers or coefficients.

• The base \( e \) signifies exponential growth and decay.
• The exponent \( t \) can either increase or decrease the function rapidly.
• Negative exponents like \( e^{-t} \) indicate decay, representing the reciprocal \( \frac{1}{e^t} \).

Understanding these properties helps in differentiating and solving equations involving exponential terms.

Derivative Calculation

Calculating the derivative of exponential functions follows specific rules. The derivative of \( e^t \) is simply \( e^t \), while for \( e^{-t} \), it's \(-e^{-t} \). In our scenario, the function \( F(t) = Ue^t + Ve^{-t} \) requires differentiation.

• The term \( Ue^t \) differentiates to \( Ue^t \).
• The term \( Ve^{-t} \) differentiates to \(-Ve^{-t} \).

Thus, the derivative of the function becomes \( F'(t) = Ue^t - Ve^{-t} \).It's imperative to correctly identify and compute these derivatives, as they help us find critical points by setting \( F'(t) = 0 \). This process unveils where the rate of change of the function is zero, signaling a potential maximum, minimum, or saddle point in the graph of the function.

Solving Equations

Solving equations often involves setting expressions equal and isolating variables. In our exercise, we solve \( F'(t) = 0 \) to find critical points. This involves several steps:1. Start with setting \( Ue^t - Ve^{-t} = 0 \).2. Rearrange to \( Ue^t = Ve^{-t} \).3. Multiply by \( e^t \) to eliminate negative exponent: \( Ue^{2t} = V \).Next, we solve for \( e^{2t} \) by expressing it as \( e^{2t} = \frac{V}{U} \).

• Taking the natural logarithm of both sides, \( \ln(e^{2t}) = \ln\left(\frac{V}{U}\right) \), leads to solving for \( 2t \).
• Use the property \( \ln(e^x) = x \) resulting in \( 2t = \ln\left(\frac{V}{U}\right) \).
• Isolating \( t \), we divide by 2: \( t = \frac{1}{2} \ln\left(\frac{V}{U}\right) \).

This \( t \) value is then verified in the original derivative to confirm it truly is a critical point, emphasizing the vital role of solving equations in calculus.