qz02sol

# qz02sol - r = &amp;amp;lt; 5 4sin , 4cos , 4sin...

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Math 53 Section 002 Quiz II Solution 1. Suppose that u , v , w are space vectors and u · ( v × w ) = 3. Find ( a ) ( u × v ) · w ( b ) v · ( u × w ) ( c ) ( u × v ) · v Solution. ( a ) and ( b ) use the properties on page 818 (Theorem 8). ( u × v ) · w = u · ( v × w ) (by 5. ) = 3 . v · ( u × w ) = v · ( w × u ) (by 1.) = ( v × w ) · u (by 5.) = u · ( v × w ) = 3 . For ( c ) , since u × v is orthogonal to v (also u ), we get ( u × v ) · v = 0 . 2. Identify the surface whose equation is given by ρ = 2cos ϕ . Solution. Multiply both sides by ρ , we get ρ 2 = 2 ρ cos ϕ . Recall that we have ρ 2 = x 2 + y 2 + z 2 and z = ρ cos ϕ in spherical coordinates. So we get x 2 + y 2 + z 2 = 2 z , x 2 + y 2 +( z 1 ) 2 = 1 . So the surface is the sphere with center ( 0 , 0 , 1 ) and radius 1. 3. Find a vector function that representations the curve of intersection of the cylin- der y 2 + z 2 = 16 and the plane x + z = 5. Then find parametric equations for the tangent line to this curve at the point ( 1 , 0 , 4 ) . Solution. Since y 2 + z 2 = 16, we may let y = 4cos θ and z = 4sin θ . Then we get x = 5 z = 5 4sin θ . So a vector function of the curve is given by

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Unformatted text preview: r = &lt; 5 4sin , 4cos , 4sin &gt;, 2 . To get the equation for the tangent line to the curve at the point ( 1 , , 4 ) , we first find the parameter for that points. Set up the equation ( 1 , , 4 ) = ( 5 4sin , 4cos , 4sin ) , we get = / 2. Now we have r = &lt; 4cos , 4sin , 4cos &gt;, r ( / 2 ) = &lt; 4cos 2 , 4sin 2 , 4cos 2 &gt; = &lt; , 4 , &gt; . 1 Hence &lt; , 4 , &gt; is a tangent vector to the curve at ( 1 , , 4 ) . So an equation of the tangent line is given by x = 1 , y = 4 t , z = 4 . Extra Credit. The moves to identify the two knots are given as follows 2...
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## qz02sol - r = &amp;amp;lt; 5 4sin , 4cos , 4sin...

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