## mathstina 3 years ago Show that the sum of the x, y, z intercepts of any tangent plane to the surface √x+√y+√z= √c is a constant.

1. TuringTest

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

i ve get the eqn for tangent plane?

3. TuringTest

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

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

5. TuringTest

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

6. TuringTest

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

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

8. mathstina

how to simplify?

9. TuringTest

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

ok.

11. TuringTest

$\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

hw to get the second eqn?

13. TuringTest

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

14. mathstina

ok!

15. TuringTest

$\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

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

17. mathstina

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

18. TuringTest

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

19. TuringTest

$\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

k. hw do i sub for each intercept?

21. mathstina

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

22. TuringTest

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

23. TuringTest

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

24. mathstina

im stuck

25. TuringTest

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

26. mathstina

r −1/2 z=p 1/2 +q 1/2 +r 1/2

27. TuringTest

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

28. mathstina

z= 1

29. TuringTest

how do you get that?

30. mathstina

r −1/2 z=r 1/2

31. TuringTest

where did p and q go?

32. TuringTest

besides, what you wrote would imply that z=r

33. mathstina

as x and y is 0

34. mathstina

show me an eg

35. TuringTest

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

36. TuringTest

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

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

38. TuringTest

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

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

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

41. mathstina

how to solve for the unknowns?

42. TuringTest

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

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

44. mathstina

hw to get the sum?

45. TuringTest

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

46. TuringTest

47. mathstina

eg?

48. TuringTest

$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

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

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

51. mathstina

sry nt sure

52. TuringTest

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

53. mathstina

wow hw do do this?

54. TuringTest

to do what?

55. mathstina

simplify hw?

56. TuringTest

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

57. TuringTest

do you see what to do from here at all?

58. mathstina

no.pls

59. TuringTest

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

60. TuringTest

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

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

62. TuringTest

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

$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

ok .thanks a lot!!!!!

65. TuringTest

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