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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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agentx5Best ResponseYou've already chosen the best response.1
Fixed the typo with the fraction command, press F5 to refresh and make it display properly. Am I understanding this question right @TuringTest ?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
It's looking correct to me...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
except that you can plug in the value for t... which just makes y=x+7
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
Wait what? You mean just plug in 1?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
why not, you need to know the value of the slope at that point, no?
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
2/2 = 1 and it goes away
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yeah, is that right? do you know?
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
lol that was it /facepalm at myself >_<
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
haha sweet, it happens :)
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
Well everybody got to see my work then lol
 one year ago

mykoBest ResponseYou've already chosen the best response.0
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
 one year ago

agentx5Best ResponseYou've already chosen the best response.1
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
 one year ago

mykoBest ResponseYou've already chosen the best response.0
my way is shorter, :). You can avoid step 2. But ya, good job
 one year ago

mykoBest ResponseYou've already chosen the best response.0
no need of that, becouse actually sometimes it will not work, if the dy/dx is infinity for example
 one year ago

mykoBest ResponseYou've already chosen the best response.0
just keep y and x same level variables without considering one of them like the independent one till the end
 one year ago
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