# Homework Help: Curvature and radius of curvature of a cartesian equation

1. Sep 17, 2015

### TyroneTheDino

1. The problem statement, all variables and given/known data
A highway has an exit ramp that beings at the origin of a coordinate system and follows the curve
$y=\frac{1}{32}x^{\frac{5}{2}}$ to the point (4,1). Then it take on a circular path whose curvature is that given bt the curve $y=\frac{1}{32}x^{\frac{5}{2}}$ at the point (4,1). If the curve is smooth at (4,1), (that is differentiable at (4,1)), find the equation of the circle who arc fomrs the path after (4,1)

2. Relevant equations
$R= \frac{(1+(y')^{2})^{3/2}}{y''}$

3. The attempt at a solution
I basically used the equation for radius of curvature to find what the radius of the circle would be.

First I derive the original equation to get: y'=$\frac{5}{64}x^{\frac{2}{3}}$

and then y''= $\frac{10}{192}x^{\frac{-1}{3}}$

I then put these into the Radius of Curvature equation and got:

$R= \frac{(1+(\frac{5}{64}x^{\frac{2}{3}})^{2})^{3/2}}{\frac{10}{192}x^{\frac{-1}{3}}}$

The value that they said they were viewing the curve at was x=4. So I evaluate the equation I made at x=4.
The value I end up with is 32.24 for the radius.

This is where I run into trouble. Trying to find the equation of this circle perplexes me.

I realize because the point we looked at was x=4 so I should shift the circle right that many units. I know that the circle should be tangent to the line $y=\frac{1}{32}x^{\frac{5}{2}}$. So I believe it should be shift up 1+R. I come up with the equation (x-4)^2+(y-32.24)^2=1039.4176. Graphing this with the original line gives something that looks nothing like what I expect. the circle is definitely not tangent to the curve so I am wondering where I went wrong.

2. Sep 17, 2015

### SteamKing

Staff Emeritus
You should double check this derivative.

3. Sep 17, 2015

### Ray Vickson

How do you go from $y = c_1 x^{5/2}$ to $y' = c_2 x^{2/3}$? When did $(5/2) - 1$ become $2/3$?

Also: your formula for R can give a negative result for some curves, so you should have $|y^{\prime \prime}|$ in the denominator (although in this particular case it makes no difference).

Last edited: Sep 17, 2015
4. Sep 17, 2015

### Staff: Mentor

Very minor point: In English, we differentiate something to get its derivative.

5. Sep 17, 2015

### TyroneTheDino

Sorry that was a very silly mistake when I differentiated incorrectly.

$y'=\frac{5}{64}x^{\frac{3}{2}}$

$y''=\frac{15}{128}x^{\frac{1}{2}}$

So now

$\frac{((1+\frac{5}{64}x^{\frac{3}{2}})^2)^(3/2)}{\frac{15}{128}x^{\frac{1}{2}}}$

Evaluating this at x=4 makes R=6.99

R^2=48.98

I create the equation (x+4)^2+(y-6.99)^2=48.68

Now that I found the proper derivative the circle equation still appears to be wrong.

6. Sep 17, 2015

### RUber

Try finding the center of the circle by using the radius and the fact that the slope of your curve $y=\frac{1}{32}x^{\frac{5}{2}}$ at the point (4,1) is tangent to the circle. It will require a little geometry.

7. Sep 17, 2015

### SteamKing

Staff Emeritus
Now, your problem appears to be correctly evaluating R when x = 4. I don't get R = 6.99 when x = 4.

8. Sep 17, 2015

### TyroneTheDino

So I kind of took what you said and reasoned with it. I noticed that center of the circle in the y direction would just be the radius so I just said (y-6.99)^2 for that part. Now for how much to shift the circle to the right or left I plugged in the value 4 for x and 1 for y so that I could see what value the x part could be shifted over. I got the equation (x-.3897)^2+(y-6.99)^2=48.86 as my result. Does that make sense?

Last edited: Sep 17, 2015
9. Sep 17, 2015

### SteamKing

Staff Emeritus
You might have missed my previous post. Check your arithmetic when calculating R for x = 4. I don't get R = 6.99.

10. Sep 17, 2015

### TyroneTheDino

I think this confusion comes from my lack of perentheses. My fault again. Or not what result are you getting? I check and still get the same thing

$\frac{\left (1+\left (\frac{5}{64}x^{\frac{3}{2}} \right )^{2} \right )^{3/2}}{\frac{15}{128}x^{\frac{1}{2}}}$

Sorry if the notation is confusing.

Last edited: Sep 17, 2015
11. Sep 17, 2015

### TyroneTheDino

Am I allowed to repost a question if I feel like I didn't get the help I needed?

12. Sep 17, 2015

### SteamKing

Staff Emeritus
What sort of additional help do you think you are missing?

13. Sep 17, 2015

### Staff: Mentor

No, don't start a new thread with the same problem. If you're not getting the help you need, just say that, along with some explanation of what you're having trouble with.

14. Sep 17, 2015

### TyroneTheDino

Ok, thank you...

I am not very confident with my reasoning, I feel like it could be right, but at the same time worried that I am making things up.

Using the curvature equation I found that the radius of stated circle in the problem would be $\frac{89\sqrt{89}}{120}$.

Using this radius I begin to create the equation for the circle. I decided that the circle would need to be shifted up the amount the radius is. I start with $(x-?)^2+(y-\frac{89\sqrt{89}}{120})^2= \frac{704969 }{14400}$

Now, I need to solve for what x is shifted left or right by. I know that the circle will be tangent to the point (4,1). I plugged these values into the circle equation I have so far providing:
$(4-?)^2+(1-\frac{89\sqrt{89}}{120})^2= \frac{704969 }{14400}$
$(4-?)^2=\frac{7921}{576}$
$(4-?)=\frac{89}{24}$
?=7/24

Using this shift I determine my final equation for the circle to be:
$(x-\frac{7}{24})^2+(y-\frac{89\sqrt{89}}{120})^2= \frac{704969 }{14400}$

Is the rationale behind my logic understandable? Or is there something I missed?

15. Sep 17, 2015

### Ray Vickson

I don't get exactly $R = 6.99$ either, but I (or, rather Maple) do get $R = (89)^{3/2} /120 \doteq 6.996869340 \doteq 7.00$.

16. Sep 17, 2015

### Staff: Mentor

There's a typo in the above. It should be
$$\frac{(1+(\frac{5}{64})^2 x^3)^{3/2}}{\frac{15}{128}x^{1/2}}$$
In the numerator you had $(1 + y'^2)^{3/2}$. $y' = \frac{5}{64}x^{3/2}$, so $y'^2 = (\frac 5 {64})^2 (x^{3/2})^2 = \frac {25}{64^2} x^3$.
@SteamKing, I'm getting $R = \frac{89 \sqrt{89}}{120} \approx 6.997$ when x = 4. Is your comment about rounding or did Tyrone and I make a different error?

17. Sep 17, 2015

### RUber

I still disagree with your idea to assume that the center of the circle is one radius above or below y=1.
Your slope at (4,1) is .625. Since the problem says that your circle continues smoothly from this point, you know that the tangent line to the circle of radius 7 (or so) has this same slope. That means that the slope of the line from (4,1) to the center be -1/(.625)= -1.6 using the fact that the radius is always perpendicular to the tangent.
You can solve this triangular problem with $d_y = - 1.6 d_x$ so by the Pythagorean theorem, $(d_x)^2 + (1.6d_x)^2 = 6.997^2$ This will give you the difference in x from 4 to the center, and putting that back into the ratio from the slope will give you the shift from 1 to the center in y.
Once you have the proper center, and you have the radius, you will be able to get the correct equation for the circle.

18. Sep 17, 2015

### TyroneTheDino

I see what you mean. (I think)
So I did what you said.
$3.56dx^{2}=(\frac{89 \sqrt{89}}{120})^{2}$

$dx=\frac{89}{24}$

$4-\frac{89}{24}=\frac{7}{24}$
Using $\frac{89}{24}$. I plug into dy to distance from y=1. $dy=(\frac{89}{15})$.
$1+(\frac{89}{15})=\frac{104}{15}$

The final equation is:
$\left (x-\frac{7}{24} \right )+\left (y-\frac{104}{15} \right )=\left (\frac{\sqrt{89}89}{120} \right )^{2}$