Tangent Curve Questions

**SportfreundeKeaneKent** · May 24th 2007, 05:00 PM

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

**Jhevon** · May 24th 2007, 05:44 PM

Originally Posted by SportfreundeKeaneKent

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

First Question:

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

Thus, the curves intersect where:
$\sqrt {3 - 2x^2} = \sqrt {x}$

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

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

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

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

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

when $x = 1$ :

$y = \sqrt {1} = 1$

So $(1,1)$ 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:

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

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

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

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

For curve 2:
$x = y^2$

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

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

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

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

Question 14:

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

By the product rule:

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

$\Rightarrow \frac {dy}{dx} = -3(x - 2)^2 + 3$ 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, $y'$ is max at $(2,3)$

So we have that the tangent of greatest slope occurs at $x = 2$ and it's value is 3. I think you can take it from here and find the equation of the tangent line

**SportfreundeKeaneKent** · June 1st 2007, 05:43 PM

Originally Posted by Jhevon

First Question:

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

when $x = 1$ :

$y = \sqrt {1} = 1$

So $(1,1)$ 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.

Shouldn't (1,-1) also be a point of intersection since 1=y^2 and y= +/- 1 or am I plugging it into the wrong equation here or something?

**Jhevon** · June 1st 2007, 06:37 PM

Originally Posted by SportfreundeKeaneKent

Shouldn't (1,-1) also be a point of intersection since 1=y^2 and y= +/- 1 or am I plugging it into the wrong equation here or something?

you're absolutely right, my boy! (1,-1) is also a an intersecting point, my apologies. good looking out

**curvature** · June 2nd 2007, 07:49 AM

The two curves are perpendicular at the points of intersection.

**CaptainBlack** · June 2nd 2007, 07:56 AM

Originally Posted by curvature

The two curves are perpendicular at the points of intersection.

The diagram adds nothing to the answers and is in fact misleading as the
scale on the x and y axes are different, which alters the apparent angle
of intersection

RonL

**curvature** · June 2nd 2007, 08:02 AM

the largest slope occurs at the piont of inflection

**curvature** · June 2nd 2007, 08:06 AM

Originally Posted by CaptainBlack

The diagram adds nothing to the answers and is in fact misleading as the
scale on the x and y axes are different, which alters the apparent angle
of intersection

RonL

I just want to visualize the problem.

**curvature** · June 2nd 2007, 08:11 AM

Originally Posted by Jhevon

First Question:

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

Thus, the curves intersect where:
$\sqrt {3 - 2x^2} = \sqrt {x}$

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

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

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

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

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

when $x = 1$ :

$y = \sqrt {1} = 1$

So $(1,1)$ 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:

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

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

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

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

For curve 2:
$x = y^2$

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

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

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

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

Question 14:

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

By the product rule:

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

$\Rightarrow \frac {dy}{dx} = -3(x - 2)^2 + 3$ 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, $y'$ is max at $(2,3)$

So we have that the tangent of greatest slope occurs at $x = 2$ and it's value is 3. I think you can take it from here and find the equation of the tangent line

No, $y'$ is max at $(2,2).$

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