Question

\(x=a \cos t+\frac{b}{2} \cos 2 t, \quad y=a \sin t+\frac{b}{2} \sin 2 t\) $\frac{d x}{d t}=-a \sin t-b \sin 2 t, \quad \frac{d y}{d t}=a \cos t+b \cos 2 t$ \(\frac{d y}{d x}=\frac{-(a \cos t+b \cos 2 t)}{a \sin t+b \sin 2 t}\) $\frac{d^{2} y}{d x^{2}}=\left[\begin{array}{l}\frac{(a \sin t+b \sin 2 t)(+a \sin t+2 b \sin 2 t)}{(a \sin t+b \sin 2 t)^{2}} \\ +(a \cos t+b \cos 2 t)(a \cos t+2 b \cos 2 t)\end{array}\right] \times \frac{1}{-(a \sin t+b \sin 2 t)}$ For \(\frac{d^{2} y}{d x^{2}}=0\) \((a \sin t+b \sin 2 t)(a \sin t+2 b \sin 2 t)+(a \cos t+b \cos 2 t)\) \((a \cos t+2 b \cos 2 t)=0\) \(\Rightarrow a^{2}+2 b^{2}+a b \cos t+2 a b \cos t=0\) \(\Rightarrow \cos t=\frac{-\left(a^{2}+2 b^{2}\right)}{3 a b}\)

Short answer

Answer: The second derivative of y with respect to x will be equal to zero when \(\cos{t} = \frac{-(a^{2}+2b^{2})}{3ab}\).

Step-by-step solution

Find the first derivative of y with respect to x

We are given that \(\frac{dx}{dt} = -a\sin{t} - b\sin{2t}\) and \(\frac{dy}{dt} = a\cos{t} + b\cos{2t}\). We need to find \(\frac{dy}{dx}\) using the chain rule. Recall that \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) are given, therefore: \(\frac{dy}{dx} = \frac{a\cos{t} + b\cos{2t}}{-a\sin{t} - b\sin{2t}}\)

Find the second derivative of y with respect to x

Now, we need to find \(\frac{d^{2}y}{dx^{2}}\). We will apply the chain rule again, keeping in mind that: \(\frac{d^{2}y}{dx^{2}} = \frac{d(\frac{dy}{dx})}{dt} \cdot \frac{dt}{dx}\) First, let's find \(\frac{d(\frac{dy}{dx})}{dt}\) by differentiating \(\frac{dy}{dx}\) with respect to t: \(\frac{d(\frac{dy}{dx})}{dt} = \left[\frac{(a\sin{t} + b\sin{2t})(a\sin{t} + 2b\sin{2t})}{(a\sin{t} + b\sin{2t})^{2}} + (a\cos{t} + b\cos{2t})(a\cos{t} + 2b\cos{2t})\right] \cdot \frac{1}{-(a\sin{t} + b\sin{2t})} \) Next, we find \(\frac{dt}{dx}\), which is the reciprocal of \(\frac{dx}{dt}\): \(\frac{dt}{dx} = \frac{1}{-a\sin{t} - b\sin{2t}}\) Now, multiply both expressions we found: \(\frac{d^{2}y}{dx^{2}} = \frac{d(\frac{dy}{dx})}{dt} \cdot \frac{dt}{dx} = \left[\frac{(a\sin{t} + b\sin{2t})(a\sin{t} + 2b\sin{2t})}{(a\sin{t} + b\sin{2t})^{2}} + (a\cos{t} + b\cos{2t})(a\cos{t} + 2b\cos{2t})\right]\)

Set the second derivative equal to zero

In order to find the conditions for which the second derivative is zero, we need to set \(\frac{d^{2}y}{dx^{2}} = 0\) and solve for t: \((a\sin{t} + b\sin{2t})(a\sin{t} + 2b\sin{2t}) + (a\cos{t} + b\cos{2t})(a\cos{t} + 2b\cos{2t}) = 0\) From the given equation, we have \((a\cos{t} + 2b\cos{2t}) = 0\). Therefore, we can simplify the equation: \(a^{2} + 2b^{2} + ab\cos{t} + 2ab\cos{t} = 0\) \(\Rightarrow \cos{t} = \frac{-(a^{2}+2b^{2})}{3ab}\) Thus, under the condition \(\cos{t} = \frac{-(a^{2}+2b^{2})}{3ab}\), the second derivative of y with respect to x will be equal to zero.