VIEWS: 2 PAGES: 9 CATEGORY: Education POSTED ON: 6/21/2011
Solution Assignment # 5 Note: Text in blue color is the information related to the given question and text in black color gives the solution of the question. Q 1. Let C be the curve with parametric equations x = t , y = t2 , z = t3 ; t≥0 Find parametric equations for the tangent line to C at the point corresponding to t = 2. Solution: DEFINITION: → → If a vector-valued function r (t ) has a tangent vector r ′(t0 ) at any point t0 on its → graph, then the line parallel to r ′(t0 ) and passing through the tip of the radius → → → vector r (t0 ) is called the tangent line of the graph of r (t ) at r (t0 ) . Vector equation of the tangent line at t0 is → → → r = r (t0 ) + t r ′(t0 ) − − − − − − − − − − ( A) We will use the above definition for the solution of this question. → The vector equation r (t ) of the given curve C is → ^ ^ ^ ^ ^ ^ r (t ) = x i + y j + z k = t i + t 2 j + t 3 k for t ≥ 0 The tangent vector to C at the point corresponding to t is → ^ ^ ^ r ′(t ) = i + 2t j + 3t 2 k In particular, radius vector of the point corresponding to t = 2 is → ^ ^ ^ r (2) = 2 i + 4 j + 8 k And a tangent vector at the point whose radius vector is as above is → ^ ^ ^ r ′(2) = i + 4 j + 12 k For vector equation of tangent line at the point where t = 2, put values in Equation (A) → ^ ^ ^ → → r = x i + y j + z k = r (2) + t r ′(2) ^ ^ ^ ⎛ ^ ^ ^ ⎞ ⎛^ ^ ^ ⎞ x i + y j + z k = ⎜ 2 i + 4 j + 8 k ⎟ + t ⎜ i + 4 j + 12 k ⎟ ⎝ ⎠ ⎝ ⎠ ^ ^ ^ ^ ^ ^ = 2 i + 4 j + 8 k + t i + 4t j + 12t k ^ ^ ^ = ( 2 + t ) i + ( 4 + 4t ) j + ( 8 + 12t ) k Equating the corresponding components gives the parametric equation of tangent line at the point where t = 2 x = 2 + t , y = 4 + 4t , z = 8 + 12t → ^ ^ Q 2. Find parametric equations for the curve r (t ) = (3t − 2) i + (4t + 3) j using arc length s as a parameter. Use the point on the curve where t = 0 as the reference point. Solution: → ^ ^ ^ ^ Since r (t ) = x i + y j = (3t − 2) i + (4t + 3) j So in parametric form we can write x = 3t − 2 ⎫ ⎬ − − − − − − − − − −(1) y = 4t + 3⎭ Now rewrite these parametric equations with u in place of t x = 3u − 2 y = 4u + 3 dx =3 du dy =4 du Theorem: Let C be a curve in 2D-space given parametrically by x = x(t) , y = y (t) / / where x (t) and y (t) are continuous functions. If an arc-length parameter s is introduced with its reference point at (x(t0), y (t0)), then the parameters s and t are related by 2 2 ⎛ dx ⎞ ⎛ dy ⎞ t s=∫ ⎜ ⎟ + ⎜ ⎟ du − − − − − − − − − − − − − − − ( A) t0 ⎝ du ⎠ ⎝ du ⎠ Use this theorem for further procedure. Since it is stated in the question that use the point on the curve where t = 0 as the reference point, so here t0 = 0 . Put values in equation (A) 2 2 ⎛ dx ⎞ ⎛ dy ⎞ t s=∫ ⎜ ⎟ + ⎜ ⎟ du t0 ⎝ du ⎠ ⎝ du ⎠ t s =∫ ( 3) + ( 4 ) 2 2 du 0 t =∫ 9 + 16 du 0 t = ∫ 0 5 du =5u0 t = 5 (t − 0) = 5t s So, s = 5t ⇒ t = 5 Substituting the value of t in the parametric equations (1). ⎛s⎞ x = 3⎜ ⎟ − 2 ⎝5⎠ ⎛s⎞ y = 4⎜ ⎟ + 3 ⎝5⎠ Or 3 x = s−2 5 4 y = s+3 5 Q 3. Determine whether the following differential is exact? If so, find z by integration of exact differentials. dz = ( y sec 2 x + sec x tan x) dx + (tan x + 2 y ) dy Solution: As we know that if z = f(x, y) then ∂z ∂z dz = dx + dy ∂x ∂y Let ∂z ⎫ P= ⎪ ∂x ⎪ ∂z ⎬ − − − − − − − − − − − −( A) Q= ⎪ ∂y ⎪ ⎭ Then above differential can be written as dz = P dx + Q dy Test for exact differential: Any differential dz = P dx + Q dy − − − − − − − −( B ) where P and Q are functions of x and y, is an exact differential if ∂P ∂Q = ∂y ∂x Equate the given differential dz = ( y sec2 x + sec x tan x) dx + (tan x + 2 y ) dy with equation (B) we get P = y sec 2 x + sec x tan x and Q = tan x + 2 y ∂P = sec2 x ∂y ∂Q = sec2 x ∂x ∂P ∂Q Since = , so the given differential is exact. ∂y ∂x Integration of Exact differential to determine z: For an exact differential dz = P dx + Q dy where ∂z ⎫ P= ⎪ ∂x ⎪ ∂z ⎬ − − − − − − − − − − − − As in equation ( A) Q= ⎪ ∂y ⎪ ⎭ z = ∫ P dx + f ( y ) − − − − − − − − − − − − − (i ) where f ( y ) is an arbitrary function of y only and is same as cons tan t of int egration in a normal int egration and also z = ∫ Q dy + f ( x) − − − − − − − − − − − − − − − − − −(ii ) where f ( x) is an arbitrary function of x only and is same as cons tan t of int egration in a normal int egration Since expressions (i) and (ii) has same left hand sides so their right hand sides are also equal. Compare both expressions and find the value of f(y) and f(x), and show z. Since it is proved that the given differential is exact, so we will do the above stated procedure to determine z. P = y sec 2 x + sec x tan x and Q = tan x + 2 y z = ∫ P dx = ∫ ( y sec 2 x + sec x tan x ) dx = y tan x + sec x + f ( y ) − − − − − − − (1) z = ∫ Q dy = ∫ ( tan x + 2 y ) dy = y tan x + y 2 + f ( x) − − − − − − − − − − − −(2) Compare right hand side of both the expressions (1) and (2) we get f ( y) = y 2 and f ( x) = sec x Thus z = y tan x + sec x + y 2 Q 4. Evaluate ∫ C xy dx + x 2 dy if a. C consists of line segments from (2, 1) to (4, 1) and from ( 4, 1) to (4, 5) . b. C is the line segment from ( 2, 1) to (4, 5) . Solution: (a) Let us sub-divide C into two parts c1 (line segment joining the points (2, 1) to (4, 1) ) and c2 (line segment joining the points ( 2, 1) to (4, 5) ) , as shown in the figure. Line integral along C may be expressed as a sum of two line integrals, the first along c1 and the second along c2. That is, ∫ C xy dx + x 2 dy = ∫ xy dx + x 2 dy + ∫ xy dx + x 2 dy − − − − − − − ( A) c1 c2 c1 is parallel to line y = 1. So dy = 0. To find the limits of integral we can see that along c1 , x varies from 2 to 4. Put values in given line integral 4 ∫ c1 xy dx + x 2 dy = ∫ x (1) dx + x 2 (0) 2 4 = ∫ x dx 2 4 x2 = = 8 − 2 = 6 − − − − − − − − − − − (1) 2 2 c2 is parallel to line x = 4. So dx = 0. To find the limits of integral we can see that along c2 , y varies from 1 to 5. Put values in given line integral 5 ∫ c2 xy dx + x 2 dy = ∫ xy (0) + (4) 2 dy 1 5 = 16 ∫ dy 1 = 16 y 1 5 = 16 (5 − 1) = 16(4) = 64 − − − − − − − − − (2) Put values from equations (1) and (2) in (A) ∫ C xy dx + x 2 dy = ∫ xy dx + x 2 dy + ∫ xy dx + x 2 dy − − − − − − − ( A) c1 c2 = 6 + 64 = 70 Thus line integral along C is 70. (b) Here, since C is the line segment from ( 2, 1) to (4, 5) . So first we will find the equation of C. Let ( x1 , y1 ) = (2, 1) and ( x2 , y2 ) = (4, 5) then y2 − y1 Slope = m = x2 − x1 5 −1 4 m= = =2 4−2 2 Use Point-Slope form of the line, to find equation of C. y − y1 = m( x − x1 ) y − 1 = 2( x − 2) y = 2x − 4 +1 = 2x − 3 Now dy = 2 dx As we are converting given line integral to one variable x, so for limits of integral we see how x varies along C, which is from 2 to 4. Put values in given line integral 4 ∫ xy dx + x dy = 2 ∫ x(2 x − 3) dx + x 2 ( 2 dx ) C 2 4 =∫ ( 2x 2 − 3x + 2 x 2 ) dx 2 4 =∫ ( 4x 2 − 3 x ) dx 2 4 x3 x2 = 4 −3 3 2 2 ⎛ 256 48 ⎞ ⎛ 32 ⎞ =⎜ − ⎟ − ⎜ − 6⎟ ⎝ 3 2 ⎠ ⎝ 3 ⎠ 512 − 144 − 64 + 36 = 6 340 170 = = 6 3 Attention please: In the above question we get different values of the integral while considering two separate paths joining the same two end points. Can anyone of you give me the reason? Send me through mail.