Originally Posted by
Jhevon 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