Find the slope of the tangent to the curve of intersection of the surface 36z = 4x^2 + 9y^2 and the plane x = 3 at the point (3, 2, 2)

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Find the slope of the tangent to the curve of intersection of the surface 36z = 4x^2 + 9y^2 and the plane x = 3 at the point (3, 2, 2)

MIT 18.01 Single Variable Calculus (OCW)
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Even though this problem asks a three dimensional question, it is not really too terribly strange. Forget about the surface, and instead think about the curve generated by the intersection of the surface and the plane. Because the plane is defined by x=3, our curve will be all the points on the surface that correspond to x=3. Therfore, \[36z=4x^2+9y^2\] becomes \[36z=4(3)^2 + 9y^2\] or \[36z=36+9y^2\] Factoring out the nine and shifting around gives: \[z=1+\frac{ y^2 }{ 4 }\] Furthermore, since all points on the curve correspond to x=3, the point (3,2,2) becomes (2,2) and we have only to find the slope of this curve at this point. \[\frac{ dz }{ dy }=\frac{ y }{ 2 }\] so the slope at (2,2) is \[\frac{ dz }{ dy }=\frac{ 2 }{ 2 }=1\]

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