Question

Show that the sum of the x, y, z intercepts of any tangent plane to the surface √x+√y+√z= √c is a constant.

Answer 1

similar to the last one in approach; get the equation of the tangent plane for a point on the surface P=(p,q,r)

Answer 2

i ve get the eqn for tangent plane?

Answer 3

no, you need to find it
you need the gradient at that point (normal to the plane) and a point on the plane

Answer 4

\[1/2[1/\sqrt{p} +1/\sqrt{q}+1/\sqrt{r}]=0\]

Answer 5

yes, now dot that with a general vector in the plane to get the equation of the tangent plane

Answer 6

actually you should have written that as a vector\[\nabla f(p,q,r)=\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\]

Answer 7

\[1/2{p/\sqrt{p}(x-p) + q/\sqrt{q}(y-q) +r/\sqrt{r}(z-r)} =0\]

Answer 8

how to simplify?

Answer 9

I think you got an extra p,q and r somehow\[\nabla f(p,q,r)=\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]

Answer 10

ok.

Answer 11

\[\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]\[p^{-1/2}x+q^{-1/2}y+r^{-1/2}z=p^{1/2}+q^{1/2}+r^{1/2}\]

Answer 12

hw to get the second eqn?

Answer 13

you mean the part on the right? I just simplified the dot product and moved all the constants to one side

Answer 14

ok!

Answer 15

\[\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]\[p^{-1/2}x-p^{1/2}+q^{-1/2}y-q^{1/2}+r^{-1/2}z-r^{1/2}=0\]\[p^{-1/2}x+q^{-1/2}y+r^{-1/2}z=p^{1/2}+q^{1/2}+r^{1/2}\]

Answer 16

now plug in for each intercept
x=0, y=0, z=0
add the three results, what do you get?

Answer 17

what happened to the 1/2 for the dot product?

Answer 18

all that stuff was =0, so multiplying both sides by 2 it is still =0, but simpler

Answer 19

\[\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]\[\implies\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]

Answer 20

k. hw do i sub for each intercept?

Answer 21

is it (x, y,z) =1 respectively

Answer 22

x=0 will give you one equation, y=0 another, z=0 another
add them all up

Answer 23

sorry, I guess I mean xy=0, yz=0, xz=0

Answer 24

im stuck

Answer 25

to find the x-intercept, plug in y=0 and z=0
what do you get?

Answer 26

r −1/2 z=p 1/2 +q 1/2 +r 1/2

Answer 27

you plugged in x=0 and y=0, so this is going to be the z-intercept
solve for z

Answer 28

z= 1

Answer 29

how do you get that?

Answer 30

r −1/2 z=r 1/2

Answer 31

where did p and q go?

Answer 32

besides, what you wrote would imply that z=r

Answer 33

as x and y is 0

Answer 34

show me an eg

Answer 35

but x and y being zero does not change the value of p and q

Answer 36

p, q, and r are the coordinates of the point *on the shape* from which we got our tangent plane
x, y, and z can be anywhere in the plane, but the equation of the plane still depends on p, q, and r

Answer 37

the x-intercept *of the plane* will be found by plugging in y=z=0 to our formula *for the plane*

Answer 38

x-intercept:\[p^{-1/2}x=p^{1/2}+q^{1/2}+r^{1/2}\implies x=p+p^{1/2}q^{1/2}+q^{1/2}r^{1/2}\]now find the y and z-intercepts

Answer 39

sorry, typo\[p^{-1/2}x=p^{1/2}+q^{1/2}+r^{1/2}\implies x=p+p^{1/2}q^{1/2}+p^{1/2}r^{1/2}\]

Answer 40

q −1/2 y=p 1/2 +q 1/2 +r 1/2 ⟹y=p1/2q1q+p 1/2 r 1/2

Answer 41

how to solve for the unknowns?

Answer 42

you don't, we are trying to show that this is true for *all* points that are on the surface so we have to leave it general. same reason we went ahead with the last one in terms of a,b, and c

Answer 43

so get the z-intercept, then add the x, y, and z-intercept together

Answer 44

hw to get the sum?

Answer 45

just add them; it will look ugly at first, but a miracle will happen...

Answer 46

erm, just add the intercepts...

Answer 47

eg?

Answer 48

\[x_0=p+p^{1/2}q+p^{1/2}r^{1/2}\]\[y_0=p^{1/2}q^{1/2}+q+q^{1/2}r^{1/2}\]\[z_0=p^{1/2}r^{1/2}+q^{1/2}r^{1/2}+r\]

Answer 49

the sum of the intercepts is then\[x_0+y_0+z_0=p+q+r+2p^{1/2}q^{1/2}+2p^{1/2}r^{1/2}+2q^{1/2}r^{1/2}\]

Answer 50

this looks ugly, but it can be factored
look at the original function... what is the relationship?

Answer 51

sry nt sure

Answer 52

\[p+q+r+2p^{1/2}q^{1/2}+2p^{1/2}r^{1/2}+2q^{1/2}r^{1/2}=(\sqrt p+\sqrt q+\sqrt r)^2\]now do you see how this is connected to the original function?

Answer 53

wow hw do do this?

Answer 54

to do what?

Answer 55

simplify hw?

Answer 56

how did I know the above you mean? practice and recognizing forms

Answer 57

do you see what to do from here at all?

Answer 58

no.pls

Answer 59

remember that p, q, and r satisfy our original equation, so\[\sqrt p+\sqrt q+\sqrt r=\sqrt c\]

Answer 60

combine that with the fact that the sum of the intercepts we have shown to be\[(\sqrt p+\sqrt q+\sqrt r)^2=(\sqrt c)^2=c\]and you see that all intercepts for any tangent plane sum to exactly c
QED

Answer 61

how to get simplified to (p √ +q √ +r √ ) 2 ?

Answer 62

I did it by observation and recognizing the form, but...
look at the intercepts\[x_0=p+p^{1/2}q+p^{1/2}r^{1/2}=p^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[y_0=p^{1/2}q^{1/2}+q+q^{1/2}r^{1/2}=q^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[z_0=p^{1/2}r^{1/2}+q^{1/2}r^{1/2}+r=r^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]

Answer 63

\[x_0+y_0+z_0\]\[=p^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})+q^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})+r^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[=(p^{1/2}+q^{1/2}+r^{1/2})(p^{1/2}+q^{1/2}+r^{1/2})\]\[=(p^{1/2}+q^{1/2}+r^{1/2})^2\]

Answer 64

ok .thanks a lot!!!!!

Answer 65

welcome, this was a good mental exercise to start my day :)