Plot Z = X^2 + Y^2 Over The Interval

I need khổng lồ find all points on the surface where the normal vector passes through the point \$(0,0,2)\$.

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I started with \$ abla F = langle -2x,2y,1 angle\$

Then, since the normal vector needs to pass through \$(0,0,2)\$ & \$(x,y,z)\$ then

\$(-2x,2y,1) = lambda(x,y,z-2)\$

So I get \$lambda = pm1over 2\$

I over up getting \$z=3over 2\$ & \$z=5over 2.\$

When \$z=5over 2\$, it appears lớn overshoot the point. It also appears that valid points only occur along the line \$y=0\$, so is it safe to assume that because this came from the relationship lớn \$y\$ it can be thrown out?

Therefore I get \$x = pm sqrtfrac 32\$ and therefore \$(sqrtfrac 32,0,3over 2)\$ and \$(-sqrtfrac 32,0,3over 2)\$.

Looking at the graph I should also get the point where \$z=0\$ & I"m not sure where that point gets left out & if I"m missing any others.

Is this the correct approach?

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edited Apr 5, 2022 at 7:55

Dr. Sundar
asked Apr 5, 2022 at 7:50

lobo1335lobo1335
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At the point \$(x,y,z)\$ the gradient (the normal vector) is

\$ n = (2 x, - 2 y, -1) \$

The equation of the normal line is

\$P(t) = (x, y, z) + t (2 x, -2 y, -1) \$

Setting this equal to \$(0,0,2)\$ results in

\$ (1 + 2 t) x = 0 , (1 - 2 t) y = 0, z = 2+t\$, and in addition we have

\$ z = x^2 - y^2 \$

If \$x =0, y=0\$ then we have the point \$(0,0,0)\$.

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If \$x = 0, y e 0\$ then \$t = dfrac12,\$ và \$z = dfrac52 \$ , however, \$y^2 = x^2 - z \$ which has no real solutions.

If \$x e 0, y = 0 \$ then \$ t= -dfrac12, z = dfrac32 ,x^2 = z + y^2 \$ , for which the solutions are \$x = pm sqrtdfrac32 \$

If \$x e 0 , y e 0\$ , then there is no value of \$t\$ that will satisfy first two equations. Hence no more solutions.

Hence, there are three points \$(0,0,0), (sqrtdfrac32 , 0, dfrac32 ) , (-sqrtdfrac32 , 0, dfrac32 )\$

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answered Apr 5, 2022 at 11:06

Hosam HajeerHosam Hajeer
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A normal"s equation through \$(x,y,x^2-y^2)\$ is given by\$\$eginpmatrixx\y\x^2-y^2endpmatrix+tcdot F_x imes F_y(x,y)=eginpmatrixx\y\x^2-y^2endpmatrix+teginpmatrix-2x\2y\1endpmatrix.\$\$Equating to \$(0,0,0)\$ gives the solutions \$(x,y,t)=(0,0,-2)\$ & \$(pmsqrt3/2,0,1/2)\$.

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answered Apr 5, 2022 at 8:58

Michael HoppeMichael Hoppe