Find the slope of the tangent to the curve x = 2 tan t, y

Find the slope of the tangent to the curve $x = 2\tan t$, $y = 2\sin^{2}t + \sin 2t$ at the point $(2,2)$.

Step 1: Find the parameter value at $(2,2)$

From the first coordinate, $2\tan t = 2$, so $\tan t = 1$ and $t = \dfrac{\pi}{4}$ (taking the principal value). Checking the second coordinate confirms the choice: \[y = 2\sin^{2}\frac{\pi}{4} + \sin\frac{\pi}{2} = 2\cdot\frac12 + 1 = 2. \;\checkmark\]

Step 2: Differentiate with respect to $t$

For $y$, use $2\sin t\cos t = \sin 2t$ to tidy the first term: \[x'_{t} = \frac{2}{\cos^{2}t}, \qquad y'_{t} = 4\sin t\cos t + 2\cos 2t = 2\sin 2t + 2\cos 2t.\]

Step 3: Form the ratio

Dividing by $\dfrac{2}{\cos^{2}t}$ is multiplying by $\dfrac{\cos^{2}t}{2}$: \[\frac{dy}{dx} = \frac{2\sin 2t + 2\cos 2t}{\dfrac{2}{\cos^{2}t}} = \cos^{2}t\,\bigl(\sin 2t + \cos 2t\bigr).\]

Step 4: Evaluate at $t = \dfrac{\pi}{4}$

\[ \cos^{2}\frac{\pi}{4} = \frac12, \qquad \sin\frac{\pi}{2} = 1, \qquad \cos\frac{\pi}{2} = 0, \]

\[ k = \frac12\bigl(1+0\bigr) = \frac12. \]

Step 5: Verification

Nearby parameter values trace the curve through $(2,2)$ with the predicted slope: $t = \frac{\pi}{4}\pm 10^{-4}:\quad \frac{\Delta y}{\Delta x} \approx 0.5000 \;\;\checkmark$.

\[\text{Tangent line: } y - 2 = \tfrac12(x-2), \text{ i.e. } y = \tfrac{x}{2}+1\]