Question

parametric calculus lecture participation day 1: problem 1
(1 point)
find an equation of the tangent to the curve given by
$x = \tan(\theta)$, $y = \sec(\theta)$
at the point $(x,y)=(1,\sqrt{2})$. your answer should be in the form of $y = f(x)$ without $\theta$.
$y(x)=$
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Explanation:

Step1: Find the value of \(\theta\)

Given \(x = \tan(\theta)=1\) and \(y=\sec(\theta)=\sqrt{2}\).
We know that \(\tan(\theta)=1\) gives \(\theta=\frac{\pi}{4}+n\pi\), and \(\sec(\theta)=\sqrt{2}\) gives \(\cos(\theta)=\frac{\sqrt{2}}{2}\), so \(\theta = 2n\pi\pm\frac{\pi}{4}\).
Combining these, \(\theta=\frac{\pi}{4}\) (since the point \((1,\sqrt{2})\) is in the first - quadrant).

Step2: Calculate \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\)

We know that \(\frac{dx}{d\theta}=\sec^{2}(\theta)\) and \(\frac{dy}{d\theta}=\sec(\theta)\tan(\theta)\).
Using the formula \(\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\), we substitute the derivatives: \(\frac{dy}{dx}=\frac{\sec(\theta)\tan(\theta)}{\sec^{2}(\theta)}=\sin(\theta)\).

Step3: Evaluate \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{4}\)

When \(\theta=\frac{\pi}{4}\), \(\frac{dy}{dx}=\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\).

Step4: Use the point - slope form \(y - y_0=m(x - x_0)\)

Here \(x_0 = 1,y_0=\sqrt{2}\) and \(m=\frac{\sqrt{2}}{2}\).
\(y-\sqrt{2}=\frac{\sqrt{2}}{2}(x - 1)\)
Expand: \(y-\sqrt{2}=\frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}\)
\(y=\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}\)

Answer:

\(y=\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}\)