tangent line of a parametric curve in the form y = mx + c so for `x(t) = 5cos(2t), y(t) = t^(7/2), and t = pi/4` i need to find the tangent line at point t=pi/4 in the form y=mx+c it is my understanding that `dy/dx = (dy/dt)/(dx/dt)` and `m = dy/dx` however when i evaluate dy/dx, and find y = (Pi/4)^(7/2), my answer appears to not be correct, i double checked on this calculator (which i know is likely to be unreliable) and my logic was the same, here is the link. i only include it so you can see the working out as i dont know how to format well on this site yet and dont want to create an eyesore. I however came to the same answer, the same derivatives, and the same value for `y when t = pi/4`, and i have no idea what i am doing wrong.

$$x=5 \cos (2 t)\implies \frac{dx}{dt}=\color{red}{-}10 \sin (2 t)$$ $$y=t^{7/2}\implies \frac{dy}{dt}=\frac{7 }{2}t^{5/2}$$

$$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }=-\frac{7}{20} t^{5/2} \csc (2 t)$$ So, $$t=\frac \pi 4\implies\frac{dy}{dx}=-\frac{7 \pi ^{5/2}}{640}$$ Now, write the equation of the tangent as $$y-y_0=y'\,(x-x_0)$$ Using $x_0=0$ and $y_0=\frac{\pi ^{7/2}}{128}$ makes $$y=-\frac{7 \pi ^{5/2}}{640}\,x+\frac{\pi ^{7/2}}{128}$$ which, in decimal form, would write $$y=0.429353 -0.191334\, x$$