MHB Finding the Tangent Line to a Parametric Curve at t=\frac{\pi}{4}

[TheFallen018]

Hey guys, I've got this problem I can't seem to get past. I need to find the tangent line to a parametric curve at t=\frac{\pi}{4}

I thought I solved the equation, but my answer doesn't seem to be registered as correct. I'm guessing that means I stuffed up the equation, but I can't see where. If anyone could point me in the right direction, I would be very grateful. Here's a screenshot of my problem.

View attachment 8091

Here's the answer I came up with

$$y=-\frac{3\sqrt{\pi}}{40}x+(\frac{\pi}{4})^\frac{3}{2}$$

(My latex code doesn't seem to be working either. Is there a way I should be initialising it?)

Thanks

 

[MarkFL]

The slope of the line can be found from:

$$\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}=\frac{\dfrac{3}{2}\sqrt{t}}{-10\sin(2t)}=-\frac{3\sqrt{t}}{20\sin(2t)}$$

Hence:

$$\left.\frac{dy}{dx}\right|_{t=\frac{\pi}{4}}=-\frac{3\sqrt{\dfrac{\pi}{4}}}{20}=-\frac{3\sqrt{\pi}}{40}$$

The point on the curve is:

$$(x,y)=\left(5\cos\left(2\cdot\frac{\pi}{4}\right),\left(\frac{\pi}{4}\right)^{\Large\frac{3}{2}}\right)=\left(0,\left(\frac{\pi}{4}\right)^{\Large\frac{3}{2}}\right)$$

Thus, the line is given by:

$$y=-\frac{3\sqrt{\pi}}{40}x+\left(\frac{\pi}{4}\right)^{\Large\frac{3}{2}}\quad\checkmark$$

Your work looks correct to me.

edit: perhaps the software wants the form:

$$y=-\frac{3\sqrt{\pi}}{40}x+\frac{\pi\sqrt{\pi}}{8}$$

 

[MarkFL]

[DESMOS]advanced: {"version":5,"graph":{"viewport":{"xmin":-5.219804697224658,"ymin":-0.4416683969175965,"xmax":13.115553015424458,"ymax":5.919253619612934}},"expressions":{"list":[{"id":"graph1","type":"expression","latex":"\\left(5\\cos\\left(2t\\right),t^{\\frac{3}{2}}\\right)","domain":{"min":"0","max":"\\pi"},"color":"#2d70b3"},{"id":"3","type":"expression","latex":"\\left(0,\\left(\\frac{\\pi}{4}\\right)^{\\frac{3}{2}}\\right)","showLabel":true,"label":"Tangent Point","color":"#fa7e19","style":"POINT"},{"id":"4","type":"expression","latex":"y=-\\frac{3\\sqrt{\\pi}}{40}x+\\frac{\\pi\\sqrt{\\pi}}{8}","color":"#388c46"},{"id":"5","type":"expression","color":"#000000"}]}}[/DESMOS]

 

[TheFallen018]

The slope of the line can be found from:

$$\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}=\frac{\dfrac{3}{2}\sqrt{t}}{-10\sin(2t)}=-\frac{3\sqrt{t}}{20\sin(2t)}$$

Hence:

$$\left.\frac{dy}{dx}\right|_{t=\frac{\pi}{4}}=-\frac{3\sqrt{\dfrac{\pi}{4}}}{20}=-\frac{3\sqrt{\pi}}{40}$$

The point on the curve is:

$$(x,y)=\left(5\cos\left(2\cdot\frac{\pi}{4}\right),\left(\frac{\pi}{4}\right)^{\Large\frac{3}{2}}\right)=\left(0,\left(\frac{\pi}{4}\right)^{\Large\frac{3}{2}}\right)$$

Thus, the line is given by:

$$y=-\frac{3\sqrt{\pi}}{40}x+\left(\frac{\pi}{4}\right)^{\Large\frac{3}{2}}\quad\checkmark$$

Your work looks correct to me.

edit: perhaps the software wants the form:

$$y=-\frac{3\sqrt{\pi}}{40}x+\frac{\pi\sqrt{\pi}}{8}$$

Thank mark, I really appreciate the help. Turns out the software didn't want that form either, but that's ok. As long as I know that I've done all I can, I'm happy.

 

[MarkFL]

Thank mark, I really appreciate the help. Turns out the software didn't want that form either, but that's ok. As long as I know that I've done all I can, I'm happy.

I would speak to the professor to see what's up with that question. :)