Here's the question you clicked on:
agentx5
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?
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
Wait what? You mean just plug in 1?
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