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.

• i'd like to stress i did the working out myself first, not just took the answer from the calculator, the fact that mine is the same as the calculator makes me more worried about my logic then secure. – Darko May 16 '17 at 1:53
• MathJax tutorial here for future reference. – John Lou May 16 '17 at 1:53
• @JohnLou thank you, i intend to get familar with that soon. – Darko May 16 '17 at 1:55
• Why do you think your answer is wrong? – John Lou May 16 '17 at 2:04
• I've graphed the function and the derivative here. This looks like what you got, so I'm not sure where you made a mistake. – John Lou May 16 '17 at 2:08

$$x=5 \cos (2 t)\implies \frac{dx}{dt}=\color{red}{-}10 \sin (2 t)$$ $$y=t^{7/2}\implies \frac{dy}{dt}=\frac{7 }{2}t^{5/2}$$
$$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }=-\frac{7}{20} t^{5/2} \csc (2 t)$$ 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 $$y=0.429353 -0.191334\, x$$