- #1
tc903
- 19
- 0
Are the given families orthogonal trajectories of each other?
\(\displaystyle {x}^{2}+{y}^{2} = ax \)
\(\displaystyle {x}^{2}+{y}^{2} = by \)
I first started by finding them implicitly.
\(\displaystyle \frac{2x+a}{2y} = y' \)
\(\displaystyle \frac{2x+b}{2y} = y' \)
Then the problem wanted me to sketch my answer.
The Tschirnhausen, I solved. I just would like a better explanation.
\(\displaystyle {y}^{2} = {x}^{3}+3{x}^{2} \) Given \(\displaystyle \left(1,2\right)\) find the tangent.
I found the tangent line. \(\displaystyle \frac{9}{4}x-\frac{1}{4}\)
Then the points where there was horizontal tangent. I take the terms in the numerator and set it equal to zero. \(\displaystyle 3{x}^{2}+6x = 0 \implies 3x\left(x+2\right)=0 \implies x = 0, -2\) Then I would substitute those values in for \(\displaystyle x\) as my answers, but that is wrong, at least when \(\displaystyle x = 0\). My answer should be \(\displaystyle \left(-2,2\right) and \left(-2,-2\right)\). I was just wondering what I am missing.
\(\displaystyle {x}^{2}+{y}^{2} = ax \)
\(\displaystyle {x}^{2}+{y}^{2} = by \)
I first started by finding them implicitly.
\(\displaystyle \frac{2x+a}{2y} = y' \)
\(\displaystyle \frac{2x+b}{2y} = y' \)
Then the problem wanted me to sketch my answer.
The Tschirnhausen, I solved. I just would like a better explanation.
\(\displaystyle {y}^{2} = {x}^{3}+3{x}^{2} \) Given \(\displaystyle \left(1,2\right)\) find the tangent.
I found the tangent line. \(\displaystyle \frac{9}{4}x-\frac{1}{4}\)
Then the points where there was horizontal tangent. I take the terms in the numerator and set it equal to zero. \(\displaystyle 3{x}^{2}+6x = 0 \implies 3x\left(x+2\right)=0 \implies x = 0, -2\) Then I would substitute those values in for \(\displaystyle x\) as my answers, but that is wrong, at least when \(\displaystyle x = 0\). My answer should be \(\displaystyle \left(-2,2\right) and \left(-2,-2\right)\). I was just wondering what I am missing.
