Parametric Curves: Solving & Approximating

[misogynisticfeminist]

1. I am given a curve defined parametrically by [tex] x= 2/t , y=1-2t [/tex] i have found the equation of tangent at t=-2 to be y=4x+9, they have asked whether it cuts the curve again. how do i find that, since i don't know the original equation of the curve and can't solve them simultaneously.

2. Also they have asked to find an approximation for sec 61. I have used [tex] \frac{\delta y}{\delta x}= \frac{dy}{dx} [/tex], but i did not get the answer, 2.0605. How do i get about doing it?

 

[TD]

1. I am given a curve defined parametrically by [tex] x= 2/t , y=1-2t [/tex] i have found the equation of tangent at t=-2 to be y=4x+9, they have asked whether it cuts the curve again. how do i find that, since i don't know the original equation of the curve and can't solve them simultaneously.

Intersect the line with the curve and see if you find any other intersection points besides t = -2. Since y = 4x+9, using the parametric equation for x, y = 4(2/t) + 9. Now substitute in the parametric equation for y and solve for t.

For the second question, I'm not really sure what you mean.

 

[VietDao29]

You should first convert 61⁰ into radians (most work in calculus use radians, instead of degrees), i.e:
[tex]61 ^ \circ = \frac{61 \pi}{180} \mbox{ rad}[/tex]
Since you already have:
[tex]\sec \left( \frac{60 \pi}{180} \right) = \sec \left( \frac{\pi}{3} \right) = 2[/tex]
You can use
f(x₀ + h) ≈ f(x₀) + h f'(x₀) (h ≈ 0) to solve your problem.
Can you go from here?