If y = a sinh(ωθ) + b cosh(ωθ), where a, b, ω are constants, prove that d²y/dθ² = ω²y.
Understand the Problem
The question is asking us to prove a second derivative relationship involving a hyperbolic sine and cosine function. We will find the first and second derivatives of y with respect to theta and show that d²y/dθ² equals omega²y.
Answer
$$ \frac{d^2y}{d\theta^2} = \omega^2 y $$
Answer for screen readers
The proof shows that $$ \frac{d^2y}{d\theta^2} = \omega^2 y $$
Steps to Solve
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Define the function Let $y = \sinh(\omega \theta)$, where $\sinh$ is the hyperbolic sine function and $\omega$ is a constant.
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First Derivative To find the first derivative of $y$ with respect to $\theta$, we use the derivative of the hyperbolic sine function. The derivative is given by: $$ \frac{dy}{d\theta} = \omega \cosh(\omega \theta) $$ where $\cosh$ is the hyperbolic cosine function.
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Second Derivative Next, we find the second derivative of $y$ with respect to $\theta$. We differentiate the first derivative: $$ \frac{d^2y}{d\theta^2} = \frac{d}{d\theta}(\omega \cosh(\omega \theta)) $$ Applying the chain rule, we get: $$ \frac{d^2y}{d\theta^2} = \omega^2 \sinh(\omega \theta) $$
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Relationship Establishment Now we can relate the second derivative to $y$. Since $y = \sinh(\omega \theta)$, we substitute to find: $$ \frac{d^2y}{d\theta^2} = \omega^2 y $$
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Conclusion We have shown that: $$ \frac{d^2y}{d\theta^2} = \omega^2 y $$ Therefore, the relationship is proven.
The proof shows that $$ \frac{d^2y}{d\theta^2} = \omega^2 y $$
More Information
In this proof, we used hyperbolic functions which behave similarly to trigonometric functions but arise in the context of hyperbolas rather than circles. Hyperbolic sine and cosine functions are particularly useful in many areas of mathematics including physics and engineering.
Tips
- Misapplying derivatives: Ensure to apply the correct derivative rules for hyperbolic functions.
- Forgetting chain rule: When taking the derivative of composite functions, always remember to apply the chain rule correctly.
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