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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.
 one year ago
 one year 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.
 one year ago
 one year 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)
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

mathstina Group TitleBest ResponseYou've already chosen the best response.1
im stuck
 one year 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?
 one year 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
 one year 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
 one year ago

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

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

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

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

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

mathstina Group TitleBest ResponseYou've already chosen the best response.1
show me an eg
 one year 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
 one year 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
 one year 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*
 one year 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
 one year ago

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

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

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

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

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

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

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

mathstina Group TitleBest ResponseYou've already chosen the best response.1
sry nt sure
 one year 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?
 one year ago

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

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

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

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

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

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

mathstina Group TitleBest ResponseYou've already chosen the best response.1
how to get simplified to (p √ +q √ +r √ ) 2 ?
 one year 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\[x_0=p+p^{1/2}q+p^{1/2}r^{1/2}=p^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[y_0=p^{1/2}q^{1/2}+q+q^{1/2}r^{1/2}=q^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[z_0=p^{1/2}r^{1/2}+q^{1/2}r^{1/2}+r=r^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]
 one year ago

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

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

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