Find the equation of the tangent line to the curve at the point corresponding to the given value of the parameter:
x = t^4 + 1
y = t^3 + t
t = -1
The answer is -x, I just don't know how to go about solving it...
m (slope) = y'/x' (y-prime divided by x-prime) = (3t^2 + 1)/(4t^3) | t = -1; m = -1?
So I guess:
x = (-1)^4 + 1 = 2
y = (1)^3 + (-1) = 0
y - y1 = m(x - x1)
y - 0 = -1(x - 2)
y = -x + 2
So I don't know...either the book is wrong, or I'm wrong. I'm betting I'm wrong :)
2006-10-17
11:19:41
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4 answers
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asked by
Derek
4
in
Science & Mathematics
➔ Mathematics