Question

In Exercises 33 and \(34,\) find \(d^{2} w / d t^{2}\) using the appropriate Chain Rule. Evaluate \(d^{2} w / d t^{2}\) at the given value of \(t\) \(w=\arctan (2 x y), \quad x=\cos t, \quad y=\sin t, \quad t=0\)

Short answer

To find the second order derivative \(d^2w/dt^2\), we have to calculate the first derivative \(dw/dt\) of the function \(w = \arctan(2xy)\), substitute the given values for \(x\) and \(y\), and differentiate again to find the second derivative.

Step-by-step solution

Calculate dw/dt

We first need to calculate the first derivative of \(w\) using the chain rule. We differentiate \(w = \arctan(2xy)\) with respect to \(t\). This involves using the chain rule twice because \(x\) and \(y\) are both functions of \(t\). We get \(\frac{dw}{dt} = \frac{1}{1+(2xy)^2} \cdot (2x\frac{dy}{dt}+2y\frac{dx}{dt})\).

Substitute the derivatives of x and y

Now we replace \(\frac{dx}{dt}\) with \(-\sin t\) and \(\frac{dy}{dt}\) with \(\cos t\). After substituting these values, we get \(\frac{dw}{dt} = \frac{1}{1+(2xy)^2}(2x \cos t - 2y \sin t)\)

Compute value at t = 0

Now the first derivative has to be evaluated at \(t = 0\). Subsitute \(x = \cos0 = 1\), \(y = \sin0 = 0\) into the equation and that gives us \(\frac{dw}{dt} = 0\) at \(t = 0\)

Differentiate again to find \(d^2w/dt^2\)

Repeat steps 1-3 to calculate \(d^2w/dt^2\), i.e., differentiate \(\frac{dw}{dt}\) with respect to \(t\). Then, substitute \(t = 0\) to get the required result.