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

Mathematics
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similar to the last one in approach; get the equation of the tangent plane for a point on the surface P=(p,q,r)
i ve get the eqn for tangent plane?
no, you need to find it you need the gradient at that point (normal to the plane) and a point on the plane

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\[1/2[1/\sqrt{p} +1/\sqrt{q}+1/\sqrt{r}]=0\]
yes, now dot that with a general vector in the plane to get the equation of the tangent plane
actually you should have written that as a vector\[\nabla f(p,q,r)=\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\]
\[1/2{p/\sqrt{p}(x-p) + q/\sqrt{q}(y-q) +r/\sqrt{r}(z-r)} =0\]
how to simplify?
I think you got an extra p,q and r somehow\[\nabla f(p,q,r)=\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]
ok.
\[\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]\[p^{-1/2}x+q^{-1/2}y+r^{-1/2}z=p^{1/2}+q^{1/2}+r^{1/2}\]
hw to get the second eqn?
you mean the part on the right? I just simplified the dot product and moved all the constants to one side
ok!
\[\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]\[p^{-1/2}x-p^{1/2}+q^{-1/2}y-q^{1/2}+r^{-1/2}z-r^{1/2}=0\]\[p^{-1/2}x+q^{-1/2}y+r^{-1/2}z=p^{1/2}+q^{1/2}+r^{1/2}\]
now plug in for each intercept x=0, y=0, z=0 add the three results, what do you get?
what happened to the 1/2 for the dot product?
all that stuff was =0, so multiplying both sides by 2 it is still =0, but simpler
\[\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]\[\implies\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]
k. hw do i sub for each intercept?
is it (x, y,z) =1 respectively
x=0 will give you one equation, y=0 another, z=0 another add them all up
sorry, I guess I mean xy=0, yz=0, xz=0
im stuck
to find the x-intercept, plug in y=0 and z=0 what do you get?
r −1/2 z=p 1/2 +q 1/2 +r 1/2
you plugged in x=0 and y=0, so this is going to be the z-intercept solve for z
z= 1
how do you get that?
r −1/2 z=r 1/2
where did p and q go?
besides, what you wrote would imply that z=r
as x and y is 0
show me an eg
but x and y being zero does not change the value of p and q
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
the x-intercept *of the plane* will be found by plugging in y=z=0 to our formula *for the plane*
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
sorry, typo\[p^{-1/2}x=p^{1/2}+q^{1/2}+r^{1/2}\implies x=p+p^{1/2}q^{1/2}+p^{1/2}r^{1/2}\]
q −1/2 y=p 1/2 +q 1/2 +r 1/2 ⟹y=p1/2q1q+p 1/2 r 1/2
how to solve for the unknowns?
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
so get the z-intercept, then add the x, y, and z-intercept together
hw to get the sum?
just add them; it will look ugly at first, but a miracle will happen...
erm, just add the intercepts...
eg?
\[x_0=p+p^{1/2}q+p^{1/2}r^{1/2}\]\[y_0=p^{1/2}q^{1/2}+q+q^{1/2}r^{1/2}\]\[z_0=p^{1/2}r^{1/2}+q^{1/2}r^{1/2}+r\]
the sum of the intercepts is then\[x_0+y_0+z_0=p+q+r+2p^{1/2}q^{1/2}+2p^{1/2}r^{1/2}+2q^{1/2}r^{1/2}\]
this looks ugly, but it can be factored look at the original function... what is the relationship?
sry nt sure
\[p+q+r+2p^{1/2}q^{1/2}+2p^{1/2}r^{1/2}+2q^{1/2}r^{1/2}=(\sqrt p+\sqrt q+\sqrt r)^2\]now do you see how this is connected to the original function?
wow hw do do this?
to do what?
simplify hw?
how did I know the above you mean? practice and recognizing forms
do you see what to do from here at all?
no.pls
remember that p, q, and r satisfy our original equation, so\[\sqrt p+\sqrt q+\sqrt r=\sqrt c\]
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
how to get simplified to (p √ +q √ +r √ ) 2 ?
I did it by observation and recognizing the form, but... look at the intercepts\[x_0=p+p^{1/2}q+p^{1/2}r^{1/2}=p^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[y_0=p^{1/2}q^{1/2}+q+q^{1/2}r^{1/2}=q^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[z_0=p^{1/2}r^{1/2}+q^{1/2}r^{1/2}+r=r^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]
\[x_0+y_0+z_0\]\[=p^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})+q^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})+r^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[=(p^{1/2}+q^{1/2}+r^{1/2})(p^{1/2}+q^{1/2}+r^{1/2})\]\[=(p^{1/2}+q^{1/2}+r^{1/2})^2\]
ok .thanks a lot!!!!!
welcome, this was a good mental exercise to start my day :)

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