Parametric Curves: Solving & Approximating

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1. I am given a curve defined parametrically by x= 2/t , y=1-2t 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 \frac{\delta y}{\delta x}= \frac{dy}{dx}, but i did not get the answer, 2.0605. How do i get about doing it?
 
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misogynisticfeminist said:
1. I am given a curve defined parametrically by x= 2/t , y=1-2t 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.
 
You should first convert 610 into radians (most work in calculus use radians, instead of degrees), i.e:
61 ^ \circ = \frac{61 \pi}{180} \mbox{ rad}
Since you already have:
\sec \left( \frac{60 \pi}{180} \right) = \sec \left( \frac{\pi}{3} \right) = 2
You can use
f(x0 + h) ≈ f(x0) + h f'(x0) (h ≈ 0) to solve your problem.
Can you go from here?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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