If y = a cosh(x/a), prove that d²y/dx² = 1/a √(1 + (dy/dx)²)
Understand the Problem
The question is asking to prove a second derivative formula involving the hyperbolic cosine function. We need to differentiate the function y = a cosh(x/a) twice and then manipulate the resulting expressions to arrive at the required form d²y/dx² = 1/a √(1 + (dy/dx)²). This involves using knowledge of calculus, particularly differentiation and properties of hyperbolic functions.
Answer
$$ \frac{d^2y}{dx^2} = \frac{1}{a} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} $$
Answer for screen readers
The second derivative of $y = a \cosh\left(\frac{x}{a}\right)$ is given by
$$ \frac{d^2y}{dx^2} = \frac{1}{a} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} $$
Steps to Solve
- Differentiate the function once
We start with the function: $$ y = a \cosh\left(\frac{x}{a}\right) $$
To find the first derivative, we use the chain rule for differentiation: $$ \frac{dy}{dx} = a \cdot \sinh\left(\frac{x}{a}\right) \cdot \frac{1}{a} = \sinh\left(\frac{x}{a}\right) $$
- Differentiate the function again
Now we differentiate $ \frac{dy}{dx} $ to find the second derivative: $$ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\sinh\left(\frac{x}{a}\right)\right) $$
Applying the chain rule again, we have: $$ \frac{d^2y}{dx^2} = \cosh\left(\frac{x}{a}\right) \cdot \frac{1}{a} = \frac{1}{a}\cosh\left(\frac{x}{a}\right) $$
- Manipulate the expression for the second derivative
Now we need to manipulate our expressions to prove: $$ \frac{d^2y}{dx^2} = \frac{1}{a} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} $$
First, calculate $ \left(\frac{dy}{dx}\right)^2 $: $$ \left(\frac{dy}{dx}\right)^2 = \left(\sinh\left(\frac{x}{a}\right)\right)^2 $$
Using the hyperbolic identity: $$ \cosh^2(t) - \sinh^2(t) = 1 $$
we can write: $$ \cosh^2\left(\frac{x}{a}\right) = 1 + \sinh^2\left(\frac{x}{a}\right) $$
Thus: $$ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \sinh^2\left(\frac{x}{a}\right) = \cosh^2\left(\frac{x}{a}\right) $$
- Substitute and finish the proof
Substituting back, we have: $$ \frac{d^2y}{dx^2} = \frac{1}{a}\cosh\left(\frac{x}{a}\right) $$
And since: $$ \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \cosh\left(\frac{x}{a}\right) $$
We can finally conclude: $$ \frac{d^2y}{dx^2} = \frac{1}{a} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} $$
The second derivative of $y = a \cosh\left(\frac{x}{a}\right)$ is given by
$$ \frac{d^2y}{dx^2} = \frac{1}{a} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} $$
More Information
This relation shows a fascinating connection between hyperbolic functions and their derivatives, demonstrating how second derivatives can often relate to the first derivatives in non-linear functions. Hyperbolic functions are similar to trigonometric functions and appear frequently in calculus and physics.
Tips
- Not applying the chain rule correctly when differentiating hyperbolic functions.
- Forgetting the hyperbolic identities which relate $\sinh$ and $\cosh$.
- Miscalculating the square of derivatives when manipulating the expressions.
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