4. Given ln(x/y) + y^3 - 2x = -1.
A. Find the equation of the normal line to the curve ln(x/y) + y^3 - 2x = -1 at the point (1,1). (I got -2x+3)
B. Find the equation of a tangent line to the curve y=e^(x^2) that "also" passes through the point (1,0). You can approximate your final answer with your calculator after you've shown your work. (idk how to do this)
1 answer
-
Let's rewrite things a bit to get
lnx - lny + y^3 - 2x = -1
Now,
A.
1/x - 1/y y' + 3y^2 y' - 2 = 0
y' (3y^2 - 1/y) = 2 - 1/x
y' = (2 - 1/x) / (3y^2 - 1/y)
so at (1,1), y' = (2-1)/(3-1) = 1/2
The normal there has slope -2, so
y-1 = -2(x-1)
y = -2x + 3
You are correct
B.
At any point on the curve, say, (m,e^(m^2)) the slope is 2m e^(m^2)
So now we have a point and a slope, so the equation is
y-0 = 2m e^(m^2) (x-1)
Now, we know that at x=m,
2m e^(m^2) (m-1) = e^m^2
2m(m-1) = 1
2m^2-2m-1 = 0
m = (1±√3)/2
So there are two lines:
y = (1+√3)e^(1+√3/2) (x-1)
y = (1-√3)e^(1-√3/2) (x-1)
You can see the graph of the first line and its tangency at
https://www.wolframalpha.com/input/?i=plot+y%3De%5Ex%5E2%2C+y+%3D+%281%2B%E2%88%9A3%29e%5E%281%2B%E2%88%9A3%2F2%29+%28x-1%29+for+0%3Cx%3C2
The other line is pretty hard to see, and just clutters things up if you plot it too.