OpenStudy (anonymous):

Find the equation of the tangent plane to the surface z=4x^3*y^2+2y at point (1, -2, 12)

4 years ago
OpenStudy (amistre64):

the gradient defines the normal of the plane, and you are given the anchor point for it

4 years ago
OpenStudy (anonymous):

You could find the tangent line at that point and it would help a bit more

4 years ago
OpenStudy (anonymous):

Would the equation be 48x + 14y + z = 64?

4 years ago
OpenStudy (amistre64):

lets see :)\[z=4x^3y^2+2y\] \[F(x,y,z)=4x^3y^2+2y-z\] \[F_x=12x^2y^2\]\[F_y=8x^3y+2\]\[F_z=-1\] the normal to the plane at any point is:\[<F_x,F_y,F_z>\]

4 years ago
OpenStudy (anonymous):

Oh lol do it that way...it's a lot faster that way.....I mean the way amistre64 did

4 years ago
OpenStudy (amistre64):

(1, -2, 12) \[Fx = 12(1^2)(-2)^2=48\]\[Fx = 8(1^3)(-2)+2=-14\]\[Fz=-1\] \[P_{tan}=48(x-1)-14(y+2)-(z-12)=0\]

4 years ago
OpenStudy (anonymous):

Oh, thanks so much! Just did a careless mistake simplifying at the end. Way easier than it seemed :D

4 years ago
OpenStudy (amistre64):

:) good luck

4 years ago
OpenStudy (amistre64):

lol, the trouble with calculus is all that algebra getting in the way

4 years ago
OpenStudy (anonymous):

Indeed :p

4 years ago
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