Couple of curve tangent questions. I have no idea what the 1st is asking.

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- May 24th 2007, 05:00 PMSportfreundeKeaneKentTangent Curve Questions
Couple of curve tangent questions. I have no idea what the 1st is asking.

http://img107.imageshack.us/img107/8782/814ni0.png - May 24th 2007, 05:44 PMJhevon
First Question:

Let curve 1 be: $\displaystyle 2x^2 + y^2 = 3 \Rightarrow y = \sqrt {3 - 2x^2}$

Let curve 2 be: $\displaystyle x = y^2 \Rightarrow y = \sqrt {x}$

Thus, the curves intersect where:

$\displaystyle \sqrt {3 - 2x^2} = \sqrt {x}$

$\displaystyle \Rightarrow 3 - 2x^2 = x$

$\displaystyle \Rightarrow 2x^2 + x - 3 = 0$

$\displaystyle \Rightarrow (2x + 3)(x - 1) = 0$

$\displaystyle \Rightarrow x = - \frac {3}{2} \mbox { or } x = 1$

$\displaystyle x = - \frac {3}{2}$ is extraneous, since $\displaystyle y$ has no solution for this $\displaystyle x$-value

when $\displaystyle x = 1$:

$\displaystyle y = \sqrt {1} = 1$

So $\displaystyle (1,1)$ is the ONLY point of intersection. Now let's find the slopes of the tangent lines of each curve at this point. If the slopes are perpendicular, that is, one is the negative inverse of the other, then we have that the curves are orthogonal.

For curve 1:

$\displaystyle 2x^2 + y^2 = 3$

$\displaystyle \Rightarrow 4x + 2y \frac {dy}{dx} = 0$

$\displaystyle \Rightarrow \frac {dy}{dx} = - \frac {2x}{y}$

at $\displaystyle (1,1)$, $\displaystyle \frac {dy}{dx} = -2$

For curve 2:

$\displaystyle x = y^2$

$\displaystyle \Rightarrow 1 = 2y \frac {dy}{dx}$

$\displaystyle \Rightarrow \frac {dy}{dx} = \frac {1}{2y}$

at $\displaystyle (1,1)$, $\displaystyle \frac {dy}{dx} = \frac {1}{2}$

Thus we see that the curves are orthogonal, since -2 is the negative reciprocal of $\displaystyle \frac {1}{2}$

Question 14:

$\displaystyle y = (x - 1)^2 (4 - x)$

By the product rule:

$\displaystyle \frac {dy}{dx} = 2(x - 1)(4 - x) - (x - 1)^2$

$\displaystyle \Rightarrow \frac {dy}{dx} = -3(x - 2)^2 + 3$ I skipped a lot of steps, i don't think you'll have a problem getting here.

This is a downward opening parabola, the derivative is maximum at it's vertex. That is, $\displaystyle y'$ is max at $\displaystyle (2,3)$

So we have that the tangent of greatest slope occurs at $\displaystyle x = 2$ and it's value is 3. I think you can take it from here and find the equation of the tangent line - Jun 1st 2007, 05:43 PMSportfreundeKeaneKent
- Jun 1st 2007, 06:37 PMJhevon
- Jun 2nd 2007, 07:49 AMcurvaturethe two curves
The two curves are perpendicular at the points of intersection.

- Jun 2nd 2007, 07:56 AMCaptainBlack
- Jun 2nd 2007, 08:02 AMcurvaturethe largest slope
the largest slope occurs at the piont of inflection

- Jun 2nd 2007, 08:06 AMcurvaturevisualize the problem
- Jun 2nd 2007, 08:11 AMcurvaturewrong