Parametric Curves and Tangents

Question

I am struggling with a question regard parametric curves and finding tangents to them but something is going wrong somewhere in the process and I cannot figure out why.

The question asks: consider the parametric curve $x=3\cos(2t)$ and $y=t^{7/2}$. Find the equation of the tangent at $t=\pi/4$.

So what I've tried is using the formula $y-y_1=m(x-x_1)$. So I've solved $x$ and $y$ for $t=\pi/4$. Used these as my x and y coordinates. Next I've rearranged so that $t=3\arccos(x)/2$ and then substituted that into the equation $y=t^{7/2}$

I believe this is where my process becomes dodgy. I've taken the derivative of $x$ and $y$ and then divided $y'/x'$ and used that as my slope. Then I've rearranged the point slope formula and found my equation of the tangent line and provided it in the form of $y=mx + c$ as required.

I am honest unsure oh what I am missing here because no matter what I try I cannot seem to nail this question despite it being quite simple and straightforward. I think my process is good I am not sure if it is just a simple calculation error along the way.

Any guidance would be greatly appreciated.

Thank you.

Answer 1

While your method (substituting for $t$) is correct, the problem was getting at teaching a much simpler way to do this. The point is that $$ \frac{dy}{dx} = \frac{ {\frac{dy}{dt}} }{{\frac{dx}{dt}} } = \frac {\frac72 t^{\frac52}} {-6 \sin(2t)} $$ and at $t = \frac\pi 4$ that slope is $-\frac{7}{384} \pi^{\frac52}$. This is probably much easier than the substitution method you used.

Answer 2

The gradient of the tangent to the catesian equation is given by $\frac{dy}{dx}$. Here $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$ Whence, $$\frac{dy}{dx} = \frac{\frac{7}{2}t^{\frac{5}{2}}}{-6 \sin(2t)}$$ At $t=\frac{\pi}{4}$ \begin{eqnarray} \frac{dy}{dx} &=& \frac{\frac{7}{2}\left(\frac{\pi}{4}\right)^{\frac{5}{2}}}{-6 \sin(\frac{\pi}{2})} \\ &=& \frac{\frac{7}{2}\left(\frac{\pi}{4}\right)^{\frac{5}{2}}}{-6} \\ \end{eqnarray} Now re-arrange the above line into something meaningful and apply the straight-line technique you used above.

Answer 3

You are right in using the equation $y-y_1=m(x-x_1)$, by substituting $t$ in the parametric equations for $x$ and $y$ you will have the $x_1$ and $y_1$ coefficients, the $m$ coefficient is exactly what you have done. I didn't understand why you rearranged the $x = 3cos(2t)$ equation, that's not necessary, and probably if you let that go, you'll find the answer.