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August 7th, 2017, 02:55 PM  #1 
Member Joined: Jan 2017 From: Toronto Posts: 97 Thanks: 2  Vector Field upper&lower limit question
Compute $\displaystyle \int sin(x) ds $ sin x ds along the line segment from (1, 2, 1) to (1, 2, 5). My solution: x = 1 + 2t y = 2 z = 1 + 4t $\displaystyle \int_{a}^{b} sin(2t  1) \sqrt{20} dt $ How do I determine the upper&lower limit a & b? 
August 7th, 2017, 07:31 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,300 Thanks: 664 
$t=0$ and $t=1$

August 8th, 2017, 02:10 AM  #3 
Member Joined: Jan 2017 From: Toronto Posts: 97 Thanks: 2  Is t always within the range of 0 and 1 for any vector field problems?
Last edited by zollen; August 8th, 2017 at 02:22 AM. 
August 8th, 2017, 03:39 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,576 Thanks: 667 
That depends entirely upon how you set up the parametric equations. You have x = 1 + 2t y = 2 z = 1 + 4t as the line from (1, 2, 1) to (1, 2, 5). x= 1+ 2t= 1 so 2t= 0, t= 0. y= 2 for all t. z= 1+ 4t= 1 so 4t= 0, t= 0 again. x= 1+ 2t= 1 so 2t= 2, t= 1. y= 2 for all t. z= 1+ 4t= 5 so 4t= 4, t= 1 again. Please, please, please, learn why a formula works rather than just memorizing! I suspect that you learned that parametric equations for the line from $\displaystyle (x_0, y_0, z_0)$ to $\displaystyle (x_1, y_1, z_1)$ are given by $\displaystyle x= x_0+ (x_1 x_0)t$, $\displaystyle y= y_0+ (y_1 y_0)t$, and $\displaystyle z= z_0+ (z_1 z_0)t$. Do you see why that works? If t= 0 then we have only $\displaystyle x_0$, $\displaystyle y_0$, and $\displaystyle z_0$. When t= 1 we have $\displaystyle x_0+ x_1 x_0= x_1$, $\displaystyle y_0+ y_1 y_0= y_1$, and $\displaystyle z_0+ z_1 z_0= z_1$. But you don't need to memorize a formula. We know that, since this is a straight line, we can take the parametric equations to be "linear": x= at+ b for some a and b. We choose to take t= 0 at the first point, (1, 2, 1) and t= 1 at the second point, (1, 2, 5). So when t= 0, we want x= 1= a(0)+ b, b= 1. When t= 1, we want 1= a(1)+ b= a 1 so a= 2: x= 2t 1. Similarly, taking y= at+ b, when t= 0, we want y= 2= a(0)+ b so b= 2. When t= 1, we want y= 2= a(1)+ 2 so a= 0: y= 2. Finally, taking z= at+ b, when t= 0, we want z= 1 so z= 1= a(0)+ b so b= 1. When z= 1, z= 5= a(1)+ 1 so a= 4: z= 4t+ 1. But we could as well have chosen the parameter, t, to go from, say, 1 to 2: Since this is a line, we can take the parametric equation to be linear, x= at+ b. When t= 1, I want x= 1 so I want 1= a+ b. When t= 2, I want x= 2 so 2= 2a+ b. Subtracting a+ b= 1 from 2a+ b= 2, a= 3. Then 3+ b= 1 so b= 4: x= 3a 4. Similarly we get y= 2 and z= 4t 3. Those give, for t= 1, (1, 2, 1) and, for t= 2, (1, 2, 5) as before. It's just much simpler to take the endpoints to be at t= 0 and t= 1. While parametric equations for a straight line can be linear, they don't have to be. Since there are two points we just need two coefficients in each equation. We could, for example, take x= a t^3+ b with t from 1 to 2. Then we must have x= 1= a(1)+ b and x= 1= a( 8 )+ b. Subtracting a+ b= 1 from 8a+ b= 1, we have 7a= 2 so a= 2/7. Then a+ b= 2/7+ b= 1 so b= 1 2/7= 9/7. x= (2/7)t^3 9/7 and similarly for y and z. Of course, there is no good reason to do all that extra work when we can take the equations to be linear and the t values at the endpoints the simple numbers t= 0 and t= 1. 
August 8th, 2017, 05:29 AM  #5 
Senior Member Joined: Sep 2015 From: CA Posts: 1,300 Thanks: 664  

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field, limit, question, upper, upperandlower, vector 
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