How many tangent lines pass through...

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June 3rd 2008, 09:08 PM
SportfreundeKeaneKent

How many tangent lines pass through...

Hi

Posting this because I don't know how to go about doing it (I'm sure the actual math is easy, just the process). Anyways, it asks how many tangent lines pass through the curve y = x^2 + 2x through the point (-1/2,-3). The differentiated equation is y'=2x + 2 but how exactly would you find the number of curves passing through that?

My second question's on Lemniscate's curve (the figure 8). The equation is y^2 = x^2 - x^4. I don't understand what it means when asking "find dy/dx or y' at (1/2, sqrt3/4). Would I just plug in these two coordinates for y and x after implicitly differentiating? I get y' = 1-2x^2/1-x^2 after implicitly differentiating this.
June 3rd 2008, 09:38 PM
TheEmptySet

Quote:

Originally Posted by SportfreundeKeaneKent

Hi

Posting this because I don't know how to go about doing it (I'm sure the actual math is easy, just the process). Anyways, it asks how many tangent lines pass through the curve y = x^2 + 2x through the point (-1/2,-3). The differentiated equation is y'=2x + 2 but how exactly would you find the number of curves passing through that?

Any point on the curve $y=x^2+2x$ is an ordered pair of the form $(x,x^2+2x)$ We also have another point $(-\frac{1}{2},-3)$ So now we can calculate the slope between these two points

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{x^2+2x+3}{x+\frac{1}{2}}$

Bute we know that the slope of a tangent line is equal to the derivative

$m = f'(x) \iff \frac{x^2+2x+3}{x+\frac{1}{2}}=2x+2 \iff x^2+2x+3=2x^2+2x+x+1 \implies$

$0=x^2+x-2 \iff 0=(x+2)(x-1)$ so the x=-2 and 1

So these are the x=coordinates of the tangent lines passing thought the point. So there are two of them.

I hope this helps.
(Rock)
June 3rd 2008, 09:46 PM
o_O

Quote:

Originally Posted by SportfreundeKeaneKent

My second question's on Lemniscate's curve (the figure 8). The equation is y^2 = x^2 - x^4. I don't understand what it means when asking "find dy/dx or y' at (1/2, sqrt3/4). Would I just plug in these two coordinates for y and x after implicitly differentiating? I get y' = 1-2x^2/1-x^2 after implicitly differentiating this.

Let's do an example. Find y' at (1,2) of the equation $x^{2} + y^{2} = 5$ .

Implicit differentiation gives us:
$2x + 2yy' = 0$
$2yy' = -2x$
$y' = -\frac{x}{y} = -\frac{1}{2}$

Our expression for y' is in both terms of x and y but that's perfectly fine. We have a coordinate (1,2) to evaluate at so you simply plug it in to your expression for y' to find the slope of a line tangent to that particular point.

Your expression for y' in your particular question doesn't seem right. Show us what you've done and we'll see if we can pinpoint any mistakes you've made.
June 3rd 2008, 10:32 PM
earboth

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Quote:

Originally Posted by SportfreundeKeaneKent

...

My second question's on Lemniscate's curve (the figure 8). The equation is y^2 = x^2 - x^4. I don't understand what it means when asking "find dy/dx or y' at (1/2, sqrt3/4). Would I just plug in these two coordinates for y and x after implicitly differentiating? I get y' = 1-2x^2/1-x^2 (unfortunately wrong) after implicitly differentiating this.

to #2:

Only for the records: Your equation of the derivative is wrong:

From

$2yy'=2x-4x^3~\implies~y'=\frac{x-2x^3}{\pm \sqrt{x^2-x^4}}$ ... $\implies$ ... $y'=\frac{x(1-2x^2)}{|x| \cdot (\pm \sqrt{1-x^2})}=\pm \frac{1-2x^2}{\sqrt{1-x^2}}$

All these transformations are not necessary because you know the coordinates of the tangent point. Plug in the coordinates into the first equation and calculate y':

$2 \cdot \left(\frac{\sqrt{3}}4 \right)y'=2\cdot \frac12-4\left(\frac12 \right)^3~\implies~y'=\frac{\sqrt{3}}{3}$