query_10_1mth

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course Mth 277

query_10_1I sent you an email saying that the QA associated with this section is linked to the one before it. Do I not do anything for that QA then????

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Question: Find the domain of F(t) X G(t) when F(t) = t^2 i - (t+2)j + (t-1)k and G(t) = (1/(t+2))i + (t-5)j + sqrt(t) k.

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Your solution:

F(t)XG(t)=(-t^2 -t^3/2 - 6t - 2sqrt(t) + 5)`i - [t^5/2 - (t-1)/(t+2)]`j + [t^3 - 5t^2 + (t+2) / (t+2)]`k

domain [0,inf)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: Describe the graph of G(t) = (sin t)i + (cos t)j + (4/3)k

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Your solution:

This graph is obviously a circle.

the k component tells us the location.

Therefore, the circle is located 4/3 units in z direction.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: Given F(t)= (t)i - 5(e^t)j +(t^3)k, G(t) = ti - (1/t)k and H(t) = (t*sin t)i + (e^-t)j, find H(t) dot [G(t) X F(t)]

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Your solution:

<(t*sin t)+(e^-t)+0>dot[X<(t) - 5(e^t) +(t^3)>]

<(t*sin t)+(e^-t)+0>dot<-5e^t/t, -(t^4+1), -5te^t>

-(t*sin t)5e^t/t-(t^4+1)(e^-t)+0

-sin(t)5e^t-(t^4+1)(e^-t)

@&

I suspect this is OK, but it's unreadable due to the lack of commas to separate the components of most of your vectors.

*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: Find a vector function F whose graph is the curve given by the equation x/5 = (y-3)/6 = (z+2)/4.

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Your solution:

x=5t

y=3+6t

z=-2+4t

F(t)=(5t)i+(3+6t)j+(-2+4t)k

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: Find the limit as t -> 2 of ((t^4-2)/(t-2))i + ((t^2-4)/(t^2-2t))j + ((t^2 + 3)e^(t-2))k.

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Your solution:

limit as t -> 2 of ((t^4-2)/(t-2))i+((t^2-4)/(t^2-2t))j+((t^2+3)e^(t-2))k

obv we can not just plug in with this limit since the i component is undef so we take the deriv of top and bot

4t^3/1i+((2t)/(2t-2))j+(7*1)k

32i+2j+7k

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: How many revolutions are made by the circular helix R(t) = (sin t)i + (cos t)j + (3/4)tk in a vertical distance of 12 units.

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Your solution:

Since we are talking about the vertical distance I will set 12 equal to the formula for the k component.

12=3/4t

t=16

We know that if we divide the time by 2pi we get revolutions

16/(2pi)=2.5 revolutions

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#