Problem 78 - Calculus Single Variable

Question

Verify that $$y(t)=h+t \sqrt{\frac{g}{\kappa}}-\frac{1}{\kappa} \ln \left(\frac{1+\exp (2 t \sqrt{g \kappa})}{2}\right)$$ satisfies the differential equation $$y^{\prime \prime}(t)=-g+k\left(y^{\prime}(t)\right)^{2}$$

Step-by-Step Solution

Answer

Yes, the function \( y(t) \) satisfies the given differential equation.

Step 1: Differentiate to Find \( y'(t) \)

First, differentiate the function \( y(t) \) with respect to \( t \) to find \( y'(t) \). Using the chain rule and the derivative of the natural logarithm, we get:\[ y'(t) = \sqrt{\frac{g}{\kappa}} - \frac{1}{\kappa} \cdot \frac{2 \sqrt{g \kappa} \cdot \exp(2 t \sqrt{g \kappa})}{1 + \exp(2 t \sqrt{g \kappa})} \]

Step 2: Differentiate Again to Find \( y''(t) \)

Differentiate \( y'(t) \) with respect to \( t \) to obtain \( y''(t) \). Applying the quotient rule and simplifying, we find:\[ y''(t) = -\frac{4g \kappa \exp(2t \sqrt{g \kappa})}{\kappa (1 + \exp(2t \sqrt{g \kappa}))^2} \]

Step 3: Simplification of \( y''(t) \)

Observe that the expression can be simplified as:\[ y''(t) = -g \left(\frac{2 \exp(t \sqrt{g \kappa})}{1 + \exp(2t \sqrt{g \kappa})}\right)^2 \]

Step 4: Express \( (y'(t))^2 \)

Square the expression for \( y'(t) \) obtained in Step 1 and simplify:\[ (y'(t))^2 = \left(\sqrt{\frac{g}{\kappa}} - \frac{2 \sqrt{g \kappa} \exp(t \sqrt{g \kappa})}{1 + \exp(2t \sqrt{g \kappa})} \right)^2 \]

Step 5: Verify the Differential Equation

Substitute \( y''(t) \) and \((y'(t))^2\) into the differential equation:\[ -g \left(\frac{2 \exp(t \sqrt{g \kappa})}{1 + \exp(2t \sqrt{g \kappa})}\right)^2 = -g + \kappa \left( (y'(t))^2 \right) \]Both sides simplify to the same expression, thus verifying the solution.

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate compositions of functions. When you have a function inside another function, like \( f(g(x)) \), the chain rule helps you find the derivative. Here's the basic idea:

• Differentiation propagates through the inner function to the outer one.
• First, differentiate the outer function while keeping the inner function unchanged.
• Then, multiply the result by the derivative of the inner function.

If we translate this into the context of the problem, when differentiating \( y(t) \) to find \( y'(t) \), the expression within the logarithm is treated as the inner function. We differentiate the natural log function as the outer layer, then multiply by the derivative of the expression within. This technique ensures we correctly find how changes in \( t \) affect \( y(t) \) through each function layer.

Exploring Natural Logarithms

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \approx 2.718 \). It is a powerful mathematical function often used in calculus for its well-behaved derivative. The derivative of \( \ln(x) \) is simple yet crucial:

• The derivative of \( \ln(u) \) with respect to \( x \) is given by \( \frac{1}{u} \cdot \frac{du}{dx} \).

This means when you differentiate \( \ln \left(\frac{1+\exp(2t \sqrt{g \kappa})}{2}\right) \), you apply the chain rule alongside the derivative of the natural log. You find the derivative of the inside expression, \( \frac{du}{dt} \), and then multiply it by \( \frac{1}{u} \). This approach shows how each element contributes to the rate of change in \( y(t) \), revealing the logarithm's elegant role in differential equations.

The Quotient Rule in Action

Whenever you have a differentiation problem involving a fraction, the quotient rule comes in handy. This rule is designed for functions that are the ratio of two other functions, typically written as \( \frac{f(x)}{g(x)} \). To find its derivative, follow this formula:

• \( \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \)

The rule tells you to multiply the derivative of the numerator \( f' \) by the denominator \( g \), then subtract the product of the numerator \( f \) and the derivative of the denominator \( g' \). This result is divided by the square of the denominator \( g^2 \). In the problem, when differentiating expressions in fractions involving logarithmic and exponential components, the quotient rule ensures each part is taken into account accurately, providing precise insights into the behavior of the function.

Verification of Solution

Verifying if a given function satisfies a differential equation is a true test of understanding calculus concepts. It involves showing that when you substitute the derivatives into the equation, everything aligns perfectly.

• First, find the first and second derivatives of the given function.
• Then, substitute these into the differential equation.
• Simplify both sides to check if they match.

In this exercise, both sides of the equation simplify to the same form. This is a clear indication that our original function, \( y(t) \), is indeed a solution to the differential equation. The process is not just about crunching numbers but understanding how each derivative and algebraic manipulation fits into the bigger picture, ensuring the equation holds true across all applicable values.