Find the planes tangent to the following surfaces at the indicated points: a) x2+2y2+3xz=10, at the point (1,2,13) b) y2−x2=3, at the point (1,2,8) c) xyz=1, at the point (1,1,1)

Question

Find the planes tangent to the following surfaces at the indicated points:  a) x2+2y2+3xz=10, at the point (1,2,13)  b) y2−x2=3, at the point (1,2,8)  c) xyz=1, at the point (1,1,1)

Laith Petty · Accepted Answer

Let f:Rn→R be a differentiable function. Recal that the tangent plane of the surface consisting of points such that f(x)=k for some constant k at the point x0 is given with:  ▽f(x0)⋅(x−x0)=0 (1)  a) Here we set f(x,y,z)=x2+2y2+3xz,x0=(1,2,13) and k=10. Note that f(x0)=k.  Now we can calculate:  ▽f(x)=(2x+3z,4y,3x)⇒▽f(1,2,13)=(3,8,3)  Using (1) we easily get:  0=(3,8,3)⋅(x−1,y−2,z−13)=3(x−1)+8(y−2)+3(z−13)  Thus, the tangent plane equation is:  3x+8y+3z=20  b) Here we set f(x,y,z)=y2−x2,x0=(1,2,8) and k=3. Note that f(x0)=k  Now we can calculate:  ▽f(x)=(−2x,2y,0)⇒▽f(1,2,8)=(−2,4,0)  Using (1) we easily get:  0=(−2,4,0)⋅(x−1,y−2,z−8)=−2(x−1)+4(y−2)+0(z−8)  Thus, the tangent plane equation is:  c) Here we set f(x,y,z)=xyz,x0=(1,1,1) and k=1. Note that f(x0)=k. Now we can calculate:  ▽f(x)=(yz,xz,xy)⇒▽f(1,1,1)=(1,1,1)  Using (1) we easily get:  0=(1,1,1)⋅(x−1,y−1,z−1)=1(x−1)+1(y−1)+1(z−1)  Thus, the tangent plane equation is:  x+y+z=3

Find the planes tangent to the following surfaces at the indicated points: a) x^2+2y^2+3xz=10, at the point (1,2,\frac{1}{3}) b) y^2-x^2=3, at the point (1,2,8) c) xyz=1, at the point (1,1,1)

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