Tangent to the curve from a parametric equation

Printable View

September 26th 2012, 06:28 AM
yorkey

Tangent to the curve from a parametric equation

Hello,

I've got this question I've been labouring over.

Quote:

The parametric equations of a curve are : x = 1 + 2sin^2Ɵ, y = 4tanƟ.

From a previous question I found dy/dx to be 1 / (sinƟcos^3Ɵ).

Now it says

Quote:

"Find the equation of the tangent to the curve at the point where Ɵ = π/4, giving your answer in the form y = mx + c.

So what I did was first try get a value for x and y by plugging in π/4 in the parametric equations.

x = 1 + (sin/4)^2
x = 1.00

y = 4tan(π/4)
y = 0.05

Then I put in π/4 into the dy/dx equation.

1 / (sinπ/4)((cosπ/4)^3)
= 72.95

Using the point-slope rule:

y - y1 = m(x - x1)
y - 0.05 = 72.95(x - 1)
y = 72.95x - 72.95 + 0.05
y = 72.95x + 73

According to the mark scheme, that is WAY off, as their answer is y = 4x - 4. Where did I go wrong? Sorry if this is in the wrong category.
September 26th 2012, 07:01 AM
johnsomeone

Re: Tangent to the curve from a parametric equation

You made several computational mistakes. Were you using a calculator perhaps? This problem should be solved without one.

When $\theta = \pi/4$ :
$x(\theta) = x(\pi/4) = 1+2\sin^2(\pi/4) = 1 + 2 \left(\frac{\sqrt{2}}{2}\right)^2 = 1 + 2\left(\frac{2}{4}\right) = 1 + 1 = 2$ .

$y(\theta) = y(\pi/4) = 4\tan(\pi/4) = 4(1) = 4$ .

$\frac{dy}{dx}(\theta) = \frac{dy}{dx}(\pi/4) = \frac{1}{(\sin(\pi/4)\cos^3(\pi/4)} = \frac{1}{(\sqrt{2}/2)(\sqrt{2}/2)^3}$ .

$= \frac{1}{(\sqrt{2}/2)^4} = \frac{1}{(\sqrt{16}/16)} = \frac{1}{(4/16)} = \frac{1}{(1/4)} = 4$ .

So when $\theta = \pi/4$ , it's at the point $(2, 4)$ and has tangent line of slope $4$ .
September 26th 2012, 07:32 AM
yorkey

Re: Tangent to the curve from a parametric equation

Wow, you lost me.

How did you get from $2\sin^2(\pi/4)$ to $2 \left(\frac{\sqrt{2}}{2}\right)^2$ ?
September 26th 2012, 07:33 AM
MarkFL

Re: Tangent to the curve from a parametric equation

Another approach to find the slope of the tangent line:

$\frac{dy}{dx}=\frac{dy}{d\theta}\cdot\frac{d\theta }{dx}=\left(4\sec^2(\theta) \right)\left(\frac{1}{2\sin(2\theta)} \right)=\frac{2\sec^2(\theta)}{\sin(2\theta)}$

Hence:

$\frac{dy}{dx}\left|_{\theta=\frac{\pi}{4}}=\frac{4 }{1}=4$

We could also eliminate the parameter $\theta$ , to find the equivalent Cartesian equation:

$x=1+2\sin^2\left(\tan^{-1}\left(\frac{y}{4} \right) \right)=1+\frac{2y^2}{y^2+16}$

$xy^2+16x=3y^2+16$

Implicitly differentiate with respect to $x$ :

$x\cdot2y\cdot\frac{dy}{dx}+y^2+16=6y\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{y^2+16}{2y(3-x)}$

When $\theta=\frac{\pi}{4}$ we have $(x,y)=(2,4)$ hence, at this point, we have:

$\frac{dy}{dx}=\frac{4^2+16}{2\cdot4(3-2)}=\frac{32}{8}=4$
September 26th 2012, 11:36 AM
johnsomeone

Re: Tangent to the curve from a parametric equation

Quote:

Originally Posted by yorkey

Wow, you lost me.

How did you get from $2\sin^2(\pi/4)$ to $2 \left(\frac{\sqrt{2}}{2}\right)^2$ ?

It seems you don't know that $\sin(\pi/4) = \frac{\sqrt{2}}{2}$ .

If that's so, then you don't understand some pretty basic trigonometry. If you're taking a calculus class, then that's going to cause you serious problems. I'd suggest you take measures - self-study, a tutor, whatever - to get caught up with trigonometry.
September 26th 2012, 12:23 PM
yorkey

Re: Tangent to the curve from a parametric equation

The trigonometric rules in my textbook I know and understand. They are:

y = sinx
dy/dx = cosx

y = cosx
dy/dx = -sinx

y = tanx
dy/dx = sec^2x

But it seems that they've thrown me something they haven't taught me. I'm just asking what is that rule you're talking about. Is there a more general law for it?
September 26th 2012, 02:22 PM
skeeter

Re: Tangent to the curve from a parametric equation

Quote:

Originally Posted by yorkey

The trigonometric rules in my textbook I know and understand. They are:

y = sinx
dy/dx = cosx

y = cosx
dy/dx = -sinx

y = tanx
dy/dx = sec^2x

But it seems that they've thrown me something they haven't taught me. I'm just asking what is that rule you're talking about. Is there a more general law for it?

it's called the unit circle ... you're expected to have been exposed to it prior to enrolling in a calculus course.

http://www.snow.edu/jonathanb/images/unitcircle1.gif
September 27th 2012, 01:15 PM
yorkey

Re: Tangent to the curve from a parametric equation

Right, I've got my work cut out for myself. Thanks for the help, everyone. :-)