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Unformatted text preview: HOMEWORK 05 Solutions (1) Suggested Problems (2) Suggested Problems (3) (a) Since v1 is parallel to the xz plane, its j component will be zero. Also, the ratio of the k component to the i component will be fx (P0 ) in order to have this as its slope. v1 =i +fx (P0 )k satisﬁes these conditions. For v2 , the i component will be zero for it to be parallel to the yz plane. The ratio of the k component to the j component will be fy (P0 ). Hence, we can use v2 =j +fy (P0 )k. (b) To ﬁnd n we take the cross product of v1 and v2 : i j k n=v1 ×v2 = 1 0 fx (P0 ) = −fx (P0 )i − fy (P0 )j + k 0 1 f y ( P0 ) (c) From our vector n and point P0 (x0 , y0 , z0 ), we can write the equation of our plane as: −fx (P0 )(x − x0 ) − fy (P0 )(y − y0 )+(z − z0 ) = 0. In standard form, this is: fx (P0 )x + fy (P0 )y − z = fx (P0 )x0 + fy (P0 )y0 − z0 . (d) First, we determine where S intersects the x, y , and z axes. To ﬁnd the point P1 where S intersects the z axis, set x and y equal to zero in our equation: z = −4 → P1 (0, 0, −4). Similarly, to ﬁnd where S√ inter2 sects √ x axis, set y and z equal to zero: 0 = 2x − 4, x = ± 2 → the P2 (± 2, 0, 0). Finally, for the y axis, 0 = 4y 2 − 4, y = ±1 → P3 (0, ±1, 0). Next, we ﬁnd the tangent planes at these points using our equation derived in part (c). Taking the partial derivatives of z with respect to x and y , we ﬁnd: ∂z fx (x, y ) = ∂ x = 4x ∂z fy (x, y ) = ∂ y = 8y
1 2 HOMEWORK 05 For the plane at P1 , fx (P1 ) = 0, fy (P1 ) = 0. Plugging into our equation from part (c), we get: z = −4. √ For √ planes at P2 , fx (P2 ) = ±4 2, fy (P2 ) = 0. Plugging in, we get: the ±(4 2)x − z = 8. For the planes at P3 , fx (P3 ) = 0, fy (P3 ) = ±8. Plugging in, we get: ±(8)y − z = 8. (4) (a) At the ”top of the hill”, the tangent plane should be parallel to the xy plane. The is analogous to the two dimensional case where the tangent line is parallel to the x axis where maxima occur. If the tangent plane is parallel to the xy plane, the normal vector n should be pointed in the positive z direction. (b) If n is pointed in the positive z direction, its i and j components should be zero, and its k component should be positive. (c) In 3(b), we found that n= −fx (P0 )i −fy (P0 )j +k. If P0 is the point at the ”top of the hill”, the i and j components of n are zero and we have fx (P0 ) = fy (P0 ) = 0. (d) To look for the ”highest spot” on S , we ﬁnd the values of x and y that will make fx and fy simultaneously zero. Calculating these partial derivatives, we ﬁnd: ∂z fx = ∂ x = √ −4(x−3) 2 2 fy =
∂z ∂y =√ These will both be zero when x = 3 and y = 5. Plugging these values into our equation for z gives us z = f (3, 5) = 5. 9−(x−3)2 −(y −5)2 9−(x−3) −(y −5) −4(y −5) HOMEWORK 05 3 x 6 5
2 0 5 4 4 z 3
4 2 2 0 z 3 4 2 y 6 2 4 x 2 4 8 6
y 6 8 Figure 1. Plot of z = 2 + plane at (3, 5, 5) 9 − (x − 3)2 − (y − 5)2 and its tangent ...
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 Fall '07
 ADAMNORRIS

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