Tangent line to $$$x{\left(t \right)} = 5 \cos{\left(2 t \right)}$$$, $$$y{\left(t \right)} = t^{\frac{7}{2}}$$$ at $$$t = \frac{\pi}{4}$$$
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Calculate the tangent line to $$$x{\left(t \right)} = 5 \cos{\left(2 t \right)}$$$, $$$y{\left(t \right)} = t^{\frac{7}{2}}$$$ at $$$t = \frac{\pi}{4}$$$.
Solution
Find the values of the $$$x$$$ and $$$y$$$ coordinates, that correspond to the given $$$t_{0} = \frac{\pi}{4}$$$:
$$$x_{0} = x{\left(\frac{\pi}{4} \right)} = 0$$$
$$$y_{0} = y{\left(\frac{\pi}{4} \right)} = \frac{\pi^{\frac{7}{2}}}{128}$$$
The slope of the tangent line at $$$t = t_{0}$$$ is the derivative of the function, evaluated at $$$t = t_{0}$$$: $$$M{\left(t_{0} \right)} = \frac{dy}{dx}|_{\left(t = t_{0}\right)}$$$.
The derivative of a parametric function is given by $$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$$.
To calculate it, find the derivative of $$$x$$$ with respect to $$$t$$$: $$$\frac{dx}{dt} = - 10 \sin{\left(2 t \right)}$$$ (for steps, see derivative calculator).
Next, find the derivative of $$$y$$$ with respect to $$$t$$$: $$$\frac{dy}{dt} = \frac{7 t^{\frac{5}{2}}}{2}$$$ (for steps, see derivative calculator).
Thus, $$$\frac{dy}{dx} = - \frac{7 t^{\frac{5}{2}}}{20 \sin{\left(2 t \right)}}$$$.
Hence, $$$M{\left(t_{0} \right)} = \frac{dy}{dx}|_{\left(t = t_{0}\right)} = - \frac{7 t_{0}^{\frac{5}{2}}}{20 \sin{\left(2 t_{0} \right)}}$$$.
Next, find the slope at the given point.
$$$m = M{\left(\frac{\pi}{4} \right)} = - \frac{7 \pi^{\frac{5}{2}}}{640}$$$
Finally, the equation of the tangent line is $$$y - y_{0} = m \left(x - x_{0}\right)$$$.
Plugging the found values, we get that $$$y - \frac{\pi^{\frac{7}{2}}}{128} = - \frac{7 \pi^{\frac{5}{2}}}{640} \left(x - 0\right)$$$.
Or, more simply: $$$y = - \frac{7 \pi^{\frac{5}{2}} x}{640} + \frac{\pi^{\frac{7}{2}}}{128}$$$.
Answer
The equation of the tangent line is $$$y = - \frac{7 \pi^{\frac{5}{2}} x}{640} + \frac{\pi^{\frac{7}{2}}}{128}\approx 0.42935308206437 - 0.191334262958397 x.$$$A