Find (d³y)/(dx³) for x = a cos t, y = b sin t

Find $\dfrac{d^{3}y}{dx^{3}}$ for $x = a\cos t,\ y = b\sin t$.

Step 1: The first two derivatives

From Problem 310, \[\dot x = -a\sin t, \qquad y'_{x} = -\frac{b}{a}\cot t, \qquad y''_{xx} = -\frac{b}{a^{2}\sin^{3}t}.\]

Step 2: Differentiate $y''_{xx}$ with respect to $t$

Write it as $-\dfrac{b}{a^{2}}\bigl(\sin t\bigr)^{-3}$ and use the power rule: \[\bigl(y''_{xx}\bigr)'_{t} = -\frac{b}{a^{2}}\cdot\bigl(-3\bigr)\bigl(\sin t\bigr)^{-4}\cos t = \frac{3b\cos t}{a^{2}\sin^{4}t}.\]

Step 3: Divide by $\dot x$

\[ y'''_{xxx} = \frac{3b\cos t/\bigl(a^{2}\sin^{4}t\bigr)}{-a\sin t} = -\frac{3b\cos t}{a^{3}\sin^{5}t}. \]

Step 4: Verification from the implicit form

With $y'_{x} = -\dfrac{b^{2}x}{a^{2}y}$ and $y''_{xx} = -\dfrac{b^{4}}{a^{2}y^{3}}$, one more differentiation in $x$ gives

\[y'''_{xxx} = \frac{3b^{4}}{a^{2}y^{4}}\,y'_{x} = \frac{3b^{4}}{a^{2}y^{4}}\left(-\frac{b^{2}x}{a^{2}y}\right) = -\frac{3b^{6}x}{a^{4}y^{5}}\]

\[\text{Substituting } x = a\cos t,\ y = b\sin t:\quad -\frac{3b^{6}\,a\cos t}{a^{4}b^{5}\sin^{5}t} = -\frac{3b\cos t}{a^{3}\sin^{5}t} \;\;\checkmark\]

\[t = \frac{\pi}{2}:\ \cos t = 0 \;\Longrightarrow\; y'''_{xxx} = 0, \text{ as symmetry about the } y\text{-axis requires} \;\;\checkmark\]