
Lines Tangent to Curves
Hey, these are some multiple choice questions I can't figure out. Please feel free to answer all or selectively answer some of them. Thank you!
If m1 is the slope of the curve xy=2 and m2 is the slope of the curve x^2+y^2=3, then the point of intersection of the two curves is:
a) m1=  m2 b) m1*m2= 1 c) m1= m2 d) m1*m2 = 1
e) m1*m2 = 2

The tangent to the curve y^2 xy +9 = 0 is vertical when:
a) y=0 b) y= + or  the square root of 3 c) y= .5 d) y= + or  3
e) none of these

The equation of the tangent to the curve y= xsinx at the point (pi/2, pi/2) is
a) y= x pi b) y= .5 c) y= pi x d) y=x+ .5 e) y= x

For #2, we could solve for y and differentiate, but let's do it implicitly.
$\displaystyle y^{2}xy+9=0$..............[1]
$\displaystyle 2yy'xy'y=0$
Divide by y' and get:
$\displaystyle 2yx\frac{y}{y'}=0$
Now, take the limit: $\displaystyle \lim_{y'\rightarrow{\infty}}\left[2yx\frac{y}{y'}\right]=2yx$
$\displaystyle 2yx=0, \;\ y=\frac{x}{2}$
Sub into [1]:
$\displaystyle (\frac{x}{2})^{2}x(\frac{x}{2})+9=0$
$\displaystyle \frac{x^{2}}{4}\frac{x^{2}}{2}+9=0$
Solve for x and we get $\displaystyle x=\pm{6}$
That leaves $\displaystyle y=\pm{3}$
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Let's do it using 'regular' differentiation and see if we get the same thing.
Solve $\displaystyle y^{2}xy+9=0$ for y and get:
$\displaystyle y=\frac{\sqrt{x^{2}36}+x}{2}, \;\ y=\frac{x\sqrt{36x^{2}}}{2}$
Differentiate: $\displaystyle y'=\frac{x}{2\sqrt{x^{2}36}}+\frac{1}{2}$
Set the denominator to 0 and solve for x:
$\displaystyle 2\sqrt{36x^{2}}=0$
We can see that $\displaystyle x=\pm{6}$
Plugging this into y, we see $\displaystyle \pm{3}$
I showed you this so you may know how to find a vertical tangent in the future. Study both methods. They may come in handy.