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.

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 :

.

.

.

.

So when , it's at the point and has tangent line of slope .

Re: Tangent to the curve from a parametric equation

Wow, you lost me.

How did you get from to ?

Re: Tangent to the curve from a parametric equation

Another approach to find the slope of the tangent line:

Hence:

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

Implicitly differentiate with respect to :

When we have hence, at this point, we have:

Re: Tangent to the curve from a parametric equation

Quote:

Originally Posted by

**yorkey** Wow, you lost me.

How did you get from

to

?

It seems you don't know that .

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.

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?

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

Re: Tangent to the curve from a parametric equation

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