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 .
Hello,
I've got this question I've been labouring over.
From a previous question I found dy/dx to be 1 / (sinƟcos^3Ɵ).The parametric equations of a curve are : x = 1 + 2sin^2Ɵ, y = 4tanƟ.
Now it says
So what I did was first try get a value for x and y by plugging in π/4 in the parametric equations."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.
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.
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 .
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:
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.
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?