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?

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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?

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Fixed the typo with the fraction command, press F5 to refresh and make it display properly. Am I understanding this question right @TuringTest ?
It's looking correct to me...
except that you can plug in the value for t... which just makes y=x+7

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Wait what? You mean just plug in 1?
Ooooh!
why not, you need to know the value of the slope at that point, no?
2/2 = 1 and it goes away
yeah, is that right? do you know?
lol that was it /facepalm at myself >_<
haha sweet, it happens :)
Well everybody got to see my work then lol
i can't really read well your solution, but it is done like this: find point value for t=1, which is (0,7) find tanget vector at t=1 which is (2,2) write the equation: (x,y) = (0,7)+v(2,2) from here you can write it in the form y=something by eliminating v
Ty myko :-) Yeah that's basically my steps too I'm doing... 1. find dy & dx with respect to dt (derivative of both) 2. find dy/dx 3. find the point for when t is as given, the x & y coords 4. put in the form of y=mx+b, solve for b using the known x & y 5. rewrite y=mx+b
my way is shorter, :). You can avoid step 2. But ya, good job
no need of that, becouse actually sometimes it will not work, if the dy/dx is infinity for example
just keep y and x same level variables without considering one of them like the independent one till the end

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