## 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

1. TuringTest Group Title

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. mathstina Group Title

i ve get the eqn for tangent plane?

3. TuringTest Group Title

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

4. mathstina Group Title

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

5. TuringTest Group Title

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

6. TuringTest Group Title

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

7. mathstina Group Title

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

8. mathstina Group Title

how to simplify?

9. TuringTest Group Title

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

10. mathstina Group Title

ok.

11. TuringTest Group Title

$\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}$

12. mathstina Group Title

hw to get the second eqn?

13. TuringTest Group Title

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

14. mathstina Group Title

ok!

15. TuringTest Group Title

$\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}$

16. TuringTest Group Title

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

17. mathstina Group Title

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

18. TuringTest Group Title

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

19. TuringTest Group Title

$\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$

20. mathstina Group Title

k. hw do i sub for each intercept?

21. mathstina Group Title

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

22. TuringTest Group Title

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

23. TuringTest Group Title

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

24. mathstina Group Title

im stuck

25. TuringTest Group Title

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

26. mathstina Group Title

r −1/2 z=p 1/2 +q 1/2 +r 1/2

27. TuringTest Group Title

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

28. mathstina Group Title

z= 1

29. TuringTest Group Title

how do you get that?

30. mathstina Group Title

r −1/2 z=r 1/2

31. TuringTest Group Title

where did p and q go?

32. TuringTest Group Title

besides, what you wrote would imply that z=r

33. mathstina Group Title

as x and y is 0

34. mathstina Group Title

show me an eg

35. TuringTest Group Title

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

36. TuringTest Group Title

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

37. TuringTest Group Title

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

38. TuringTest Group Title

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

39. TuringTest Group Title

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

40. mathstina Group Title

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

41. mathstina Group Title

how to solve for the unknowns?

42. TuringTest Group Title

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

43. TuringTest Group Title

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

44. mathstina Group Title

hw to get the sum?

45. TuringTest Group Title

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

46. TuringTest Group Title

47. mathstina Group Title

eg?

48. TuringTest Group Title

$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$

49. TuringTest Group Title

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

50. TuringTest Group Title

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

51. mathstina Group Title

sry nt sure

52. TuringTest Group Title

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

53. mathstina Group Title

wow hw do do this?

54. TuringTest Group Title

to do what?

55. mathstina Group Title

simplify hw?

56. TuringTest Group Title

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

57. TuringTest Group Title

do you see what to do from here at all?

58. mathstina Group Title

no.pls

59. TuringTest Group Title

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

60. TuringTest Group Title

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

61. mathstina Group Title

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

62. TuringTest Group Title

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})$

63. TuringTest Group Title

$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$

64. mathstina Group Title

ok .thanks a lot!!!!!

65. TuringTest Group Title

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