# Shortest distance between 2 curves

1. Sep 23, 2010

### jegues

1. The problem statement, all variables and given/known data
Find the shortest distance between,

$$y = x^{2} - 8x + 15$$ and,

$$2y + 7 + 2x^{2} = 0$$

2. Relevant equations

3. The attempt at a solution

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

$$y = -x^{2} - \frac{7}{2}$$

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?

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2. Sep 23, 2010

### gabbagabbahey

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

Why not start by saying that this vector is $\textbf{d}=a\textbf{i}+b\textbf{j}+c\textbf{k}$ and then solving fo the 3 unkowns $a$, $b$, and $c$? 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...

3. Sep 23, 2010

### jegues

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

$$\vec{d} \cdot \vec{v} = \vec{0}$$ Where, $$\vec{v}$$ 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!?

Last edited: Sep 23, 2010
4. Sep 23, 2010

### Office_Shredder

Staff Emeritus
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))

5. Sep 23, 2010

### jegues

So,

$$\vec{r_{1}(t)} = t^{2} - 8t +15$$

and

$$\vec{r_{2}(t)} = -t^{2} - \frac{7}{2}$$

How's that?

EDIT: Wait,

$$\vec{r_{1}(t)} = t\hat{i}+(t^{2} - 8t +15)\hat{j}$$

$$\vec{r_{2}(t)} = t\hat{i} + (-t^{2} - \frac{7}{2})\hat{j}$$

I think these 2 look better.

Okay so for my tangent vectors,

$$\vec{v_{1}(t)} = \hat{i} + (2t -8)\hat{j}$$

$$\vec{v_{2}(t)} = \hat{i} + (-2t)\hat{j}$$

So if I dot these with d,

$$\vec{v_{1}(t)} \cdot \vec{d} = a + (2t - 8)b = 0$$

$$\vec{v_{2}(t)} \cdot \vec{d} = a + -2tb = 0$$

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,

$$\vec{r_{2}(s)} = s\hat{i} + (-s^{2} - \frac{7}{2})\hat{j}$$

and,

$$\vec{v_{2}(s)} = \hat{i} + (-2s)\hat{j}$$

therefore,

$$\vec{v_{2}(s)} \cdot \vec{d} = a + -2sb = 0$$

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

Last edited: Sep 23, 2010
6. Sep 23, 2010

### jegues

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

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

I have it solved!

Last edited: Sep 23, 2010
7. Sep 24, 2010

### gabbagabbahey

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