MTH164FA10-M1d_ans

# MTH164FA10-M1d_ans - MTH 164 Multivariable Calculus 1st...

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MTH 164: Multivariable Calculus 1 st Midterm Exam ANSWERS November 1, 2010 Part A Important Formulae? κ ( t ) =
d T ds
= | r
( t ) × r
( t ) | | r
( t ) | 3 κ ( x ) = | f
( x ) | [1 + ( f
( x )) 2 ] 3 / 2 a = v
T + κ v 2 N a T = r
( t ) · r
( t ) | r
( t ) | a N = | r
( t ) × r
( t ) | | r
( t ) |

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Tuesday, October 19, 2010 1 st Midterm Exam MTH 164 (Multivariable Calculus) 1. (20 points) Consider the surface given by the equation x 2 2 y 2 + z 2 + yz = 2 (a) Find an equation of the tangent plane to the surface at the point P (2 , 1 , 1). Answer: The equation is a level set of g ( x, y ) = x 2 2 y 2 + z 2 hence the gradient of this function is normal to the surface. g =
2 x, 4 y + z, 2 z + y
and g (2 , 1 , 1) =
4 , 5 , 1
. A plane is defined by a point and its normal vector, so with X =
x, y, z
the tangent plane is defined by g · ( X (2 , 1 , 1)) = 0 or 4 x 5 y z = 4 If you use coordinates
dx, dy, dz
=
x 2 , y 1 , z + 1
relative to the coordinate system whose origin is at (2 , 1 , 1) then you get the slightly simpler equation of a plane through the this new origin 4 d 5 dy dz = 0. (b) If the point Q (2 . 1 , 0 . 9 , k ) lies near P (2 , 1 , 1) and on the surface, use linear approxi- mation to approximate the value of k . Answer: On the tangent plane we have 4( . 1) 5( . 1) dz = 0 hence . 9 = dz = z + 1 or z = . 1. Hence the z coordinate of the corresponding point on the surface will be z = . 1 + error and we are assuming that the changes are small enough so that the error term is negligible and our linear approximation is fairly accurate. Another approach is to assume that z = z ( x, y ) is a function of x and y defined implicitly by the equation above. Then using implicit di ff erentiation on the equation above we can calculate that z x (2 , 1) = 4 and z y (2 , 1) = 5 so dz = z x dx + z y dy = 4( . 1)
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