# Lines Tangent to Curves

• Mar 19th 2008, 11:56 AM
LRoseT
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

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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

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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
• Mar 19th 2008, 12:23 PM
galactus
For #2, we could solve for y and differentiate, but let's do it implicitly.

$y^{2}-xy+9=0$..............[1]

$2yy'-xy'-y=0$

Divide by y' and get:

$2y-x-\frac{y}{y'}=0$

Now, take the limit: $\lim_{y'\rightarrow{\infty}}\left[2y-x-\frac{y}{y'}\right]=2y-x$

$2y-x=0, \;\ y=\frac{x}{2}$

Sub into [1]:

$(\frac{x}{2})^{2}-x(\frac{x}{2})+9=0$

$\frac{x^{2}}{4}-\frac{x^{2}}{2}+9=0$

Solve for x and we get $x=\pm{6}$

That leaves $y=\pm{3}$

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Let's do it using 'regular' differentiation and see if we get the same thing.

Solve $y^{2}-xy+9=0$ for y and get:

$y=\frac{\sqrt{x^{2}-36}+x}{2}, \;\ y=\frac{x-\sqrt{36-x^{2}}}{2}$

Differentiate: $y'=\frac{x}{2\sqrt{x^{2}-36}}+\frac{1}{2}$

Set the denominator to 0 and solve for x:

$2\sqrt{36-x^{2}}=0$

We can see that $x=\pm{6}$

Plugging this into y, we see $\pm{3}$

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.