Exercise · ZW-TANGENT-PLANE-003 · ★☆☆ Basic
Find the tangent plane to \(z=x^2y\) at \((1,1,1)\).
Choose one option.
Show answer
- D. \(z=2 x + y - 2\)
Explanation
\(f_x=2xy=2\) and \(f_y=x^2=1\) at \((1,1)\), so \(z=1+2(x-1)+(y-1)=2x+y-2\).
Review the concept: Tangent plane
The tangent plane to the graph of a function \(f(x,y)\) at the point \((a,b,f(a,b))\) is the plane that best approximates the surface near that point and contains the tangent lines of its two coordinate sections.
\[z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\]
More practice problems: tangent plane
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- ZW-TANGENT-PLANE-002 Find the tangent plane to z=xy at the point (2,3,6).
- ZW-TANGENT-PLANE-004 Find the tangent plane to z=√x^2+y^2 at the point (3,4,5).
- ZW-TANGENT-PLANE-005 The tangent plane to z=x^2+3y at the point (1,0,1) has normal vector:
- ZW-TANGENT-PLANE-006 Use the tangent plane to z=√x^2+y^2 at (3,4) to approximate √3.03^2+3.96^2.
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- ZW-TANGENT-PLANE-009 A tangent plane to z=x^2+y^2 is parallel to the plane z=4x-2y+7. What is the point of tangency (a,b…