- Thread starter
- #1

^{2}{r- ( h ))

Answer

r= HRL/(L-R)(L+2R)

- Thread starter leprofece
- Start date

- Thread starter
- #1

Answer

r= HRL/(L-R)(L+2R)

- Admin
- #2

I would draw a diagram as follows:

From this you can get the constraint. For the objective function, we need the volume of a spherical cap:

\(\displaystyle V=\frac{\pi h}{6}\left(3a^2+h^2 \right)\)

$a$ will be the positive $x$-intercept of the circle and $h=r+k$.

Can you proceed?

- Thread starter
- #3

it may beI would draw a diagram as follows:

View attachment 2005

From this you can get the constraint. For the objective function, we need the volume of a spherical cap:

\(\displaystyle V=\frac{\pi h}{6}\left(3a^2+h^2 \right)\)

$a$ will be the positive $x$-intercept of the circle and $h=r+k$.

Can you proceed?

V= pi(r-k)/6(3a

i must derive may I Derive now?

- Admin
- #4

No...$h=r+k$, you need to find $a$ as I described (the positive $x$-intercept of the circle), and you need to find the constraint. The most straightforward way I can think of is to use the fact that a radius of the circle drawn to the point of tangency of the line segment from $(0,H)$ to $(R,0)$ is perpendicular to this segment, and so the formula for the perpendicular or shortest distance between a point and a ine may be used to obtain the value of $k$ in terms of $R$, $H$ and $r$. Another method which yields the same result, but requires more algebra is to equate the equation of the line to the circle and require the discriminant of the resulting quadratic to be zero.it may be

V= pi(r-k)/6(3a^{2}+(r-k)^{2})

i must derive may I Derive now?

The formula for the distance $d$ between a point $\left(x_0,y_0 \right)$ and a line $y=mx+b$ is:

\(\displaystyle d=\frac{\left|mx_0+b-y_0 \right|}{\sqrt{m^2+1}}\)

We know in this case that $d=r$. Can you identify $m$ and $b$?

- Thread starter
- #5

Lets seeNo...$h=r+k$, you need to find $a$ as I described (the positive $x$-intercept of the circle), and you need to find the constraint. The most straightforward way I can think of is to use the fact that a radius of the circle drawn to the point of tangency of the line segment from $(0,H)$ to $(R,0)$ is perpendicular to this segment, and so the formula for the perpendicular or shortest distance between a point and a ine may be used to obtain the value of $k$ in terms of $R$, $H$ and $r$. Another method which yields the same result, but requires more algebra is to equate the equation of the line to the circle and require the discriminant of the resulting quadratic to be zero.

The formula for the distance $d$ between a point $\left(x_0,y_0 \right)$ and a line $y=mx+b$ is:

\(\displaystyle d=\frac{\left|mx_0+b-y_0 \right|}{\sqrt{m^2+1}}\)

We know in this case that $d=r$. Can you identify $m$ and $b$?

If the points are (0,H) y (R,0)

m must be -H/R

Now y = -Hx/R +HR/R so b = H

I dont know if L is K in your solution

Introducing in your formula

r = - -Hx/R +H/(sqrt) ( - (-H/R)^2 +1)

But I dont proceed because in the book answer it is L

There must be something wrong in my solution

- Admin
- #6

- Thread starter
- #7

I dont know if my calculation was goodI would assume your book is labeling:

\(\displaystyle L=\sqrt{R^2+H^2}\)

Because I dont know how to pul this \(\displaystyle L=\sqrt{R^2+H^2}\)[/QUOTE]

in my given solving . ??????'

- Admin
- #8

You need to determine $k$. Use the formula I gave for the distance between a point and a line where:

\(\displaystyle \left(x_0,y_0 \right)=(0,k)\)

\(\displaystyle m=-\frac{H}{R}\)

\(\displaystyle b=H\)

\(\displaystyle d=r\)

Do you see why this applies?

- Thread starter
- #9

The formula for the distance d between a point (x0,y0) and a line y=mx+b is:You need to determine $k$. Use the formula I gave for the distance between a point and a line where:

\(\displaystyle \left(x_0,y_0 \right)=(0,k)\)

\(\displaystyle m=-\frac{H}{R}\)

\(\displaystyle b=H\)

\(\displaystyle d=r\)

Do you see why this applies?

d=∣∣mx0+b−y0∣∣m2+1−−−−−−√

We know in this case that d=r. Can you identify m and b?

so r= d = -H/R+H-k/(sqrt(h^2+r^2)/(R^2))

r= -H/R+H-k/((L)/(R))

So k = Lr+H/R-H/R

K = HR-Hx-rL/R

K=H-r

if I equalled and solved for r and i got nothing HR-Hx-rL/R = H-r

And now what must I do??

Derive that??)

Last edited:

- Admin
- #10

No, what you want to do is:

\(\displaystyle r=\frac{\left|-\frac{H}{R}\cdot0+H-k \right|}{\sqrt{\left(-\frac{H}{R} \right)^2+1}}\)

Now solve this for $k$ (observe that $H>k$) and substitute into the objective function. And you STILL need to find the positive $x$-intercept of the circle to use for $a$. Recall, your objective function is:

\(\displaystyle V=\frac{\pi h}{6}\left(3a^2+h^2 \right)\)

and $h=r+k$. This is why you need $k$ and $a$.

- Thread starter
- #11

con x = 0No, what you want to do is:

\(\displaystyle r=\frac{\left|-\frac{H}{R}\cdot0+H-k \right|}{\sqrt{\left(-\frac{H}{R} \right)^2+1}}\)

Now solve this for $k$ (observe that $H>k$) and substitute into the objective function. And you STILL need to find the positive $x$-intercept of the circle to use for $a$. Recall, your objective function is:

\(\displaystyle V=\frac{\pi h}{6}\left(3a^2+h^2 \right)\)

and $h=r+k$. This is why you need $k$ and $a$.

Then i got K = HR-Lr/R

and a

then??

- Admin
- #12

If you mean:

k = (HR - Lr)/R

then that is correct. However, $a$ is the positive $x$-intercept of the circle:

\(\displaystyle x^2+(y-k)^2=r^2\)

Hence:

\(\displaystyle a=\sqrt{r^2-k^2}\)

Now, put the value of $k$ in there...and try to simplify.

Now, after you do this, what do you suppose you should do with these values?

- Thread starter
- #13

2V/piH- HIf you mean:

k = (HR - Lr)/R

then that is correct. However, $a$ is the positive $x$-intercept of the circle:

\(\displaystyle x^2+(y-k)^2=r^2\)

Hence:

\(\displaystyle a=\sqrt{r^2-k^2}\)

Now, put the value of $k$ in there...and try to simplify.

Now, after you do this, what do you suppose you should do with these values?

2V/piH- H

2V/piH- H