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mathstina
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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.
 2 years ago
 2 years ago
mathstina Group Title
Show that the sum of the x, y, z intercepts of any tangent plane to the surface √x+√y+√z= √c is a constant.
 2 years ago
 2 years ago

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TuringTest Group TitleBest ResponseYou've already chosen the best response.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)
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
i ve get the eqn for tangent plane?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
no, you need to find it you need the gradient at that point (normal to the plane) and a point on the plane
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
\[1/2[1/\sqrt{p} +1/\sqrt{q}+1/\sqrt{r}]=0\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yes, now dot that with a general vector in the plane to get the equation of the tangent plane
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
actually you should have written that as a vector\[\nabla f(p,q,r)=\frac12\langle p^{1/2},q^{1/2},r^{1/2}\rangle\]
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
\[1/2{p/\sqrt{p}(xp) + q/\sqrt{q}(yq) +r/\sqrt{r}(zr)} =0\]
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
how to simplify?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I think you got an extra p,q and r somehow\[\nabla f(p,q,r)=\frac12\langle p^{1/2},q^{1/2},r^{1/2}\rangle\cdot\langle xp,yq,zr\rangle=0\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[\frac12\langle p^{1/2},q^{1/2},r^{1/2}\rangle\cdot\langle xp,yq,zr\rangle=0\]\[p^{1/2}x+q^{1/2}y+r^{1/2}z=p^{1/2}+q^{1/2}+r^{1/2}\]
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
hw to get the second eqn?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you mean the part on the right? I just simplified the dot product and moved all the constants to one side
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[\frac12\langle p^{1/2},q^{1/2},r^{1/2}\rangle\cdot\langle xp,yq,zr\rangle=0\]\[p^{1/2}xp^{1/2}+q^{1/2}yq^{1/2}+r^{1/2}zr^{1/2}=0\]\[p^{1/2}x+q^{1/2}y+r^{1/2}z=p^{1/2}+q^{1/2}+r^{1/2}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
now plug in for each intercept x=0, y=0, z=0 add the three results, what do you get?
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
what happened to the 1/2 for the dot product?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
all that stuff was =0, so multiplying both sides by 2 it is still =0, but simpler
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[\frac12\langle p^{1/2},q^{1/2},r^{1/2}\rangle\cdot\langle xp,yq,zr\rangle=0\]\[\implies\langle p^{1/2},q^{1/2},r^{1/2}\rangle\cdot\langle xp,yq,zr\rangle=0\]
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
k. hw do i sub for each intercept?
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
is it (x, y,z) =1 respectively
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
x=0 will give you one equation, y=0 another, z=0 another add them all up
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
sorry, I guess I mean xy=0, yz=0, xz=0
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
to find the xintercept, plug in y=0 and z=0 what do you get?
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
r −1/2 z=p 1/2 +q 1/2 +r 1/2
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you plugged in x=0 and y=0, so this is going to be the zintercept solve for z
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
how do you get that?
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
r −1/2 z=r 1/2
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
where did p and q go?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
besides, what you wrote would imply that z=r
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
as x and y is 0
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
show me an eg
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
but x and y being zero does not change the value of p and q
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
the xintercept *of the plane* will be found by plugging in y=z=0 to our formula *for the plane*
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
xintercept:\[p^{1/2}x=p^{1/2}+q^{1/2}+r^{1/2}\implies x=p+p^{1/2}q^{1/2}+q^{1/2}r^{1/2}\]now find the y and zintercepts
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
sorry, typo\[p^{1/2}x=p^{1/2}+q^{1/2}+r^{1/2}\implies x=p+p^{1/2}q^{1/2}+p^{1/2}r^{1/2}\]
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
q −1/2 y=p 1/2 +q 1/2 +r 1/2 ⟹y=p1/2q1q+p 1/2 r 1/2
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
how to solve for the unknowns?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
so get the zintercept, then add the x, y, and zintercept together
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
hw to get the sum?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
just add them; it will look ugly at first, but a miracle will happen...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
erm, just add the intercepts...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[x_0=p+p^{1/2}q+p^{1/2}r^{1/2}\]\[y_0=p^{1/2}q^{1/2}+q+q^{1/2}r^{1/2}\]\[z_0=p^{1/2}r^{1/2}+q^{1/2}r^{1/2}+r\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
the sum of the intercepts is then\[x_0+y_0+z_0=p+q+r+2p^{1/2}q^{1/2}+2p^{1/2}r^{1/2}+2q^{1/2}r^{1/2}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
this looks ugly, but it can be factored look at the original function... what is the relationship?
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
sry nt sure
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[p+q+r+2p^{1/2}q^{1/2}+2p^{1/2}r^{1/2}+2q^{1/2}r^{1/2}=(\sqrt p+\sqrt q+\sqrt r)^2\]now do you see how this is connected to the original function?
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
wow hw do do this?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
to do what?
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
simplify hw?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
how did I know the above you mean? practice and recognizing forms
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
do you see what to do from here at all?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
remember that p, q, and r satisfy our original equation, so\[\sqrt p+\sqrt q+\sqrt r=\sqrt c\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
how to get simplified to (p √ +q √ +r √ ) 2 ?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I did it by observation and recognizing the form, but... look at the intercepts\[x_0=p+p^{1/2}q+p^{1/2}r^{1/2}=p^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[y_0=p^{1/2}q^{1/2}+q+q^{1/2}r^{1/2}=q^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[z_0=p^{1/2}r^{1/2}+q^{1/2}r^{1/2}+r=r^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[x_0+y_0+z_0\]\[=p^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})+q^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})+r^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[=(p^{1/2}+q^{1/2}+r^{1/2})(p^{1/2}+q^{1/2}+r^{1/2})\]\[=(p^{1/2}+q^{1/2}+r^{1/2})^2\]
 2 years ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
ok .thanks a lot!!!!!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
welcome, this was a good mental exercise to start my day :)
 2 years ago
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