Contents. 9.1 Vector Functions 9.2 Motion in a Curve 9.3 Curvature and Components of Acceleration 9.4 Partial Derivatives. 9.1 Vector Functions.
Graph the curve byr(t) = 2cos ti+ 2sin tj + tk, t 0
Solutionx2 + y2 = (2cos t)2 + (2sin t)2 = 22See Fig 9.2. The curve winds upward in spiral or circular helix.
Graph the curve byr(t) = 2cos ti+ 2sin tj + 3k
Solutionx2 + y2 = (2cos t)2 + (2sin t)2 = 22, z = 3See Fig 9.3.
Find the vector functions that describes the curve Cof the intersection of y = 2x and z = 9 – x2 – y2.
SolutionLet x = t, then y = 2t, z = 9 – t2 – 4t2 = 9 – 5t2Thus, r(t) = ti+ 2tj +(9 – 5t2)k. See Fig 9.4.
Limit of a Vector Function
If exist, then
If , then (i) , c a scalar(ii) (iii)
Properties of Limits
A vector function r is said to be continuous at t = a if
(i) r(a) is defined, (ii) limta r(t)exists, and
(iii) limta r(t) = r(a).
Derivative of Vector Function
The derivative of a vector function r is (2)
for all t where the limits exists.
If , where f, g, and h are
Differentiation of Components
See Fig 9.5.
Graph the curve by r(t) = cos 2t i + sin t j, 0 t 2. Graph r’(0) and r’(/6).
Solutionx = cos 2t, y = sin t, then x = 1 – 2y2, −1 x 1andr’(t) = −2sin 2ti + cos tj,r’(0) = j, r’(/6) =
Find the tangent line to x = t2, y = t2 – t, z = −7t at t = 3
Solutionx’ = 2t, y’ = 2t – 1, z’ = −7 when t = 3,and r(3) = 9i + 6j – 21kthat is P(9, 6, –21), then we havex = 9 + 6t, y = 6 + 5t, z = –21 – 7t
If r is a differentiable vector function and s = u(t) is a
differentiable scalar function, then the derivatives ofr(s) with respect to t is
If r(s) = cos2si + sin2sj + e–3sk, s = t4, then
If r1 and r2 are differentiable vector functions and u(t)
is a differentiable scalar function.
If r(t) = 6t2i + 4e–2t j + 8cos 4t k, thenwhere c = c1i + c2j + c3k.
Consider the curve in Example 1. Since , from (3) the length from r(0) to r(t) is
Using then (4)Thus
Position vector: r(t) = t2i + tj + (5t/2)k. Graph the curve defined by r(t) and v(2), a(2).
Solutionso that See Fig 9.7.
‖v(t)‖2 = c2or v‧v = c2a(t)‧v(t) = 0
Consider the position vector in Example 2 of Sec 9.1. Graph the velocity and acceleration at t = /4.
SolutionRecall r(t) = 2cos ti + 2sin tj + 3k.then v(t) = −2sin ti + 2cos tja(t) = −2cos ti −2sin tjand
Integrating again and using r(0) = s0, Hence we have (1) See Fig 9.11
A shell is fired from ground level with v0 = 768 ft/s at an angle of elevation 30 degree. Find (a) the vector function and the parametric equations of the trajectory, (b) the maximum attitude attained, (c) the range of the shell (d) the speed of impact.
Solution(a) Initially we have s0= 0,and (2)
Since a(t) = −32j and using (2) gives (3)Integrating again,Hence the trajectory is (4)
(b) From (4), we see that dy/dt = 0 when −32t + 384 = 0 or t = 12. Thus the maximum height H is H = y(12) = – 16(12)2 + 384(12) = 2304 ft
(c) From (4) we see that y(t) = 0 when −16t(t – 24) = 0, or t = 0, 24.Then the range R is
(d) from (3), we obtain the impact speed of the shell
From (2) we have T = dr/ds, then the curvature of C
at a point is (3)
Find the curvature of a circle of radius a.
SolutionWe already know the equation of a circle isr(t) = a cos ti + a sin tj, then We getThus, (5)
The position vector r(t) = 2cos ti + 2sin tj + 3tk, find the vectors T, Nand B, and the curvature.
SolutionSincefrom (1),Next we have
Hence (7) gives N = – cos ti – sin tj
Now, Finally using and
From (4) we have
Since ||B|| = 1, it follows that (11)then (12)
The position vector r(t) = ti + ½t2j + (1/3)t3k is said to be a “twisted” cube”. Find the tangential and normal components of the acceleration at t. Find the curvature.
SolutionSince v a = t + 2t3 and From (10),
NowandFrom (11) From (12)
The level curves of f(x, y) = y2 – x2 are defined by y2 – x2 = c. See Fig 9.23. For c = 0, we obtain the lines y = x, y = −x.
Describe the level curves of F(x, y, z) = (x2 + y2)/z.
SolutionFor c 0, (x2 + y2)/z = c, or x2 + y2 = cz. See Fig 9.24. ，
If z = 4x3y2 – 4x2 + y6 + 1, find ,
If z = f(u, v) is differentiable, and u = g(x, y)and
v = h(x, y) have continuous first partial derivatives,
If z = u2 – v3, u = e2x – 3y, v = sin(x2 – y2), find and
SolutionSincethen (6) (7)
Similarly, if u1, u2,…, un are functions of a single variable t, then (10)These results can be memorized in terms of a tree diagram. See next page.
If r = x2 + y5z3and x = uve2s, y = u2 – v2s, z = sin(uvs2), find r/s.
SolutionAccording to the tree diagram,
If z = u2v3w4and u = t2, v = 5t – 8, w = t3 + t，find dz/dt.
SolutionAnother approach: differentiate z = t4(5t – 8)3(t3 + t)4