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kkurt [141]
3 years ago
9

Find the points on the surface y2 = 25 + xz that are closest to the origin.

Mathematics
1 answer:
tatuchka [14]3 years ago
5 0
You are essentially minimizing x^2+y^2+z^2 subject to y^2=25+xz. (The distance between the origin and any point (x,y,z) on the given surface is \sqrt{x^2+y^2+z^2}, but \sqrt{\mathrm{func}} and \mathrm{func} share the same critical points.)

Via Lagrange multipliers, we have Lagrangian

L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(y^2-xz-25)

with partial derivatives (set equal to 0)

L_x=2x-\lambda z=0\implies 2x=\lambda z
L_y=2y+2\lambda y=0\implies y(1+\lambda)=0
L_z=2z-\lambda x=0\implies 2z=\lambda x
L_\lambda=y^2-xz-25=0\implies y^2=xz+25

L_x=0\implies zL_x=0\implies 2xz=\lambda z^2
L_z=0\implies xL_z=0\implies 2xz=\lambda x^2
zL_x-xL_z=\lambda(z^2-x^2)=0

We assume \lambda\neq0, which means z^2=x^2\implies |z|=|x|.

L_y=0\implies y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1

In the first case, we have

y^2=25+xz\implies -xz=25

which means one of x,z must be positive, and the other is negative. From |x|=|z| we have x=-z, so

-(-z)z=z^2=25\implies z=\pm5\implies x=\mp5

So we get two critical points, (-5, 0, 5) and (5, 0, -5).

In the second case, if \lambda=-1, we get

\begin{cases}2x=-z\\2z=-x\end{cases}\implies x=z=0

which leads us to

y^2=25\implies y=\pm5

i.e. we have two additional critical points (0, 5, 0) and (0, -5, 0).

At each of these points, we get respective distances from the origin of \{5\sqrt2,5\sqrt2,5,5\}, so the two closest points to the origin on the surface y^2=25+xz are (0, 5, 0) and (0, -5, 0).
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The statements and reasons why the given statements are true are presented in the following two column proofs;

Question 1

Given: 9·(x + 6) - 41 = 75

Prove \ x = \dfrac{62}{9}

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S4. 9·x = 62                   {}R4. <u>Subtraction property</u>

S5. x = \dfrac{62}{9}                     {} R5. <u>Division property of equality</u>

Question 2.

Statement                                     {}        Reason

S1. m∠A + m∠B = m∠D   {}                     R1.  Given

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S1. ∠BDA ≅ ∠A                             {} R1. Given

S2. ∠BDA ≅ ∠CDE                      {}  R2. <u>Vertical angle theorem</u>

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Learn more here:

brainly.com/question/11331230

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