Shortest distance between 2 curves

[jegues]

Homework Statement

Find the shortest distance between,

[tex]y = x^{2} - 8x + 15[/tex] and,

[tex]2y + 7 + 2x^{2} = 0[/tex]

Homework Equations

The Attempt at a Solution

Rearranging the 2nd function into a function of y in terms of x,

[tex] y = -x^{2} - \frac{7}{2}[/tex]

From here I was able to graph the two in the x-y plane. (See figure)

Now the shortest distance between them will be a vector that runs from one curve to the other, and is perpendicular to both curves, correct?

How can I go about finding this vector? If I can somehow create this vector, I can compute his magnitude and I'll have the shortest distance between the two curves.

Any ideas?

 

[gabbagabbahey]

Now the shortest distance between them will be a vector that runs from one curve to the other, and is perpendicular to both curves, correct?

You mean perpendicular to the tangent vector of each curve, right?

How can I go about finding this vector? If I can somehow create this vector, I can compute his magnitude and I'll have the shortest distance between the two curves.

Any ideas?

Why not start by saying that this vector is [itex]\textbf{d}=a\textbf{i}+b\textbf{j}+c\textbf{k}[/itex] and then solving fo the 3 unkowns [itex]a[/itex], [itex]b[/itex], and [itex]c[/itex]? You will need 3 independent equations to solve for these variables, so try putting the condition that d is perpendicular to the tangent vector of each curve into equation form (giving you 2 equations) and then finding another equation that it must satisfy...

 

[jegues]

You mean perpendicular to the tangent vector of each curve, right?




Why not start by saying that this vector is [itex]\textbf{d}=a\textbf{i}+b\textbf{j}+c\textbf{k}[/itex] and then solving fo the 3 unkowns [itex]a[/itex], [itex]b[/itex], and [itex]c[/itex]? You will need 3 independent equations to solve for these variables, so try putting the condition that d is perpendicular to the tangent vector of each curve into equation form (giving you 2 equations) and then finding another equation that it must satisfy...

If d is to be perpendicular to the tangent vector of each curve then,

[tex] \vec{d} \cdot \vec{v} = \vec{0}[/tex] Where, [tex]\vec{v}[/tex] is the tangent vector of the given curve.

But my curves aren't in vector notation, so how do I go about finding the tangent vector?

If I can describe my two curves in the form of position vectors then I can differentiate them to find their tangent vectors, but I don't know how to describe my curves in terms of a position vector??

Can I get a few more nudges!?

 

[Office_Shredder]

Once you have solved for y as a function of x, y=f(x), then you can easily parametrize your curve to be all points (t,f(t))

 

[jegues]

Once you have solved for y as a function of x, y=f(x), then you can easily parametrize your curve to be all points (t,f(t))

So,

[tex]\vec{r_{1}(t)} = t^{2} - 8t +15[/tex]

and

[tex]\vec{r_{2}(t)} = -t^{2} - \frac{7}{2}[/tex]

How's that?

EDIT: Wait,

[tex]\vec{r_{1}(t)} = t\hat{i}+(t^{2} - 8t +15)\hat{j}[/tex]

[tex]\vec{r_{2}(t)} = t\hat{i} + (-t^{2} - \frac{7}{2})\hat{j}[/tex]

I think these 2 look better.

Okay so for my tangent vectors,

[tex]\vec{v_{1}(t)} = \hat{i} + (2t -8)\hat{j}[/tex]

[tex]\vec{v_{2}(t)} = \hat{i} + (-2t)\hat{j}[/tex]

So if I dot these with d,

[tex]\vec{v_{1}(t)} \cdot \vec{d} = a + (2t - 8)b = 0[/tex]

[tex]\vec{v_{2}(t)} \cdot \vec{d} = a + -2tb = 0[/tex]

I'm stuck with 2 equations and 3 unknowns (I don't know t!), how can I fix this!? Can I get a third equation somehow!?

EDIT: We can't assume that both curves will have the same parameter t, can we?

So,

[tex]\vec{r_{2}(s)} = s\hat{i} + (-s^{2} - \frac{7}{2})\hat{j}[/tex]

and,

[tex]\vec{v_{2}(s)} = \hat{i} + (-2s)\hat{j}[/tex]

therefore,

[tex]\vec{v_{2}(s)} \cdot \vec{d} = a + -2sb = 0[/tex]

So it looks like I need another 2 equations hmmm...

 

[jegues]

and then finding another equation that it must satisfy...

What could it be!??!? I'm so close!

EDIT: Not so close anymore had to edit post above this

I have it solved!

 

[gabbagabbahey]

So it looks like I need another 2 equations hmmm...

You also know that [itex]\textbf{d}[/itex] goes from [itex]\textbf{r}_1(t)[/itex] to [itex]\textbf{r}_2(s)[/itex], right?