• anonymous
Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x = t − t$$^{−1}$$, y = 6 + t$$^2$$, t = 1 The correct answer needs to be in the form of y=something... I have this (incorrect): $y=(\large\frac{2t^3}{t^2+1})x+7$ @ t=1, x=0 and y=7 $$\large\frac{dy}{dt}$$=2t $$\large\frac{dx}{dt}$$= 1+$$\large\frac{1}{t^2}$$ = $$\large\frac{t^2+1}{t^2}$$ Dividing by something, same as the product of its reciprocal, when doing $$\frac{dy}{dx}$$ Where's the error?
Mathematics

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