{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

M408D Quest Homework 9-solutions

# M408D Quest Homework 9-solutions - hernandez(ah29758 M408D...

This preview shows pages 1–3. Sign up to view the full content.

hernandez (ah29758) – M408D Quest Homework 9 – pascaleff – (54550) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Find lim t 0 + r ( t ) when r ( t ) = ( 4 cos t, 2 e t , 10 t ln t ) . 1. limit = ( 0 , 2 , 0 ) 2. limit = ( 4 , 0 , 10 ) 3. limit = ( 4 , 2 , 10 ) 4. limit = ( 0 , 0 , 0 ) 5. limit = ( 4 , 0 , 10 ) 6. limit = ( 4 , 2 , 0 ) correct Explanation: For a vector function r ( t ) = ( f ( t ) , g ( t ) , h ( t ) ) , the limit lim t 0 + r ( t ) = (Big lim t 0 + f ( t ) , lim t 0 + g ( t ) , lim t 0 + h ( t ) )Big . But for the given vector function, lim t 0 + f ( t ) = lim t 0 + 4 cos t = 4 , while lim t 0 + g ( t ) = lim t 0 + 2 e t = 2 , and lim t 0 + h ( t ) = lim t 0 + t ln t = 0 , using L’Hospital’s Rule. Consequently, lim t 0 + r ( t ) = ( 4 , 2 , 0 ) . keywords: vector function, limit, trig function log function, exponential function 002 10.0points A space curve is shown in black on the surface x y z Which one of the following vector functions has this space curve as its graph? 1. r ( t ) = ( t cos t, t, t sin t ) , t > 0 , 2. r ( t ) = ( sin 4 t, cos t, sin t ) 3. r ( t ) = ( t cos t, t sin t, t ) 4. r ( t ) = ( t cos t, t, t sin t ) 5. r ( t ) = ( t cos t, t sin t, t ) , t > 0 , 6. r ( t ) = ( sin 2 t, cos t, sin t ) 7. r ( t ) = ( cos t, sin t, sin 4 t ) correct 8. r ( t ) = ( cos t, sin t, sin 2 t ) Explanation:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
hernandez (ah29758) – M408D Quest Homework 9 – pascaleff – (54550) 2 The surface is a cylinder such that hori- zontal cross-sections intersect the surface in a circle. Now of the given vector functions only r ( t ) = ( cos t, sin t, sin 2 t ) r ( t ) = ( cos t, sin t, sin 4 t ) lie on such a cylinder; in particular, the sur- face is the graph of x 2 + y 2 = 1 , and the space curve passes through the point r (0) = (1 , 0 , 0) on the x -axis as well as the point r parenleftBig π 2 parenrightBig = (0 , 1 , 0) on the y -axis. But the z -coordinate of the space curve takes both positive and nega- tive values between these two points, while sin 2 t > 0 for 0 < t < π/ 2. Consequently, the space curve is the graph of r ( t ) = ( cos t, sin t, sin 4 t ) . keywords: surface, space curve, vector func- tion, 3D graph, circular cylinder, 003 10.0points Find a vector function whose graph is the curve of intersection of the circular cylinder x 2 + z 2 = 4 and the hyperbolic paraboloid y = x 2 z 2 . 1. r ( t ) = 2 cos t i + 4 sin 2 t j + 2 sin t k 2. r ( t ) = 2 sin t i 4 sin 2 t j + 2 cos t k 3. r ( t ) = 1 2 cos t i 1 4 sin 2 t j + 1 2 sin t k 4. r ( t ) = 1 2 cos t i + 1 2 cos 2 t j + 1 4 sin t k 5. r ( t ) = 2 sin t i + 4 cos 2 t j + 2 cos t k 6. r ( t ) = 2 cos t i + 4 cos 2 t j + 2 sin t k correct 7. r ( t ) = 2 sin 2 t i + 2 sin 2 t j + 2 sin 2 t k 8. r ( t ) = 2 cos 2 t i + 2 cos t j + 2 sin 2 t k Explanation: The graph of the vector function r ( t ) = x ( t ) i + y ( t ) j + z ( t ) k is the intersection of the circular cylinder x 2 + z 2 = 4 and the hyperbolic paraboloid y = x 2 z 2 when x ( t ) 2 + z ( t ) 2 = 4 , y ( t ) = 2 x ( t ) 2 z ( t ) 2 . Now cos 2 θ + sin 2 θ = 1 , which already eliminates two of the functions. On the other hand, cos 2 θ sin 2 θ = cos 2 θ , which is satisfied only by r ( t ) = 2 cos t i + 4 cos 2 t j + 2 sin t k .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 11

M408D Quest Homework 9-solutions - hernandez(ah29758 M408D...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online