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# Vector-Valued Functions Part 1 MATH 241

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Contains 12.1 to 12.3.
COURSE
Calculus III
PROF.
Dr. Roohollah Ebrahimian
TYPE
Class Notes
PAGES
6
WORDS
CONCEPTS
Math, Calculus, Derivatives, Integrals, Multivariable, vectors, functions, properties, Dot, Cross, LaTeX, Mathematics
KARMA
25 ?

## Popular in Math

This 6 page Class Notes was uploaded by Saadiq Shaik on Saturday October 1, 2016. The Class Notes belongs to MATH 241 at University of Maryland - College Park taught by Dr. Roohollah Ebrahimian in Fall 2016. Since its upload, it has received 4 views. For similar materials see Calculus III in Math at University of Maryland - College Park.

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Date Created: 10/01/16
Vector-Valued Functions Part 1 Saadiq Shaik September 2016 1 Introduction to Vector-Valued Functions A vector-valued function is function with a domain of real numbers and a range of vectors. This is in contrast to functions in prior calculus courses, which we will refer to as real-valued functions. Every VVF corresponds to 3 RVFs, referred to as component functions. F(t) = f (t)i + f (t)j + f (t)k 1 2 3 This can also be written in parametric form as x = f1(t);y = f2(t);z = 3 (t) Using the de▯nition of parametric equations given in the last chapter, we can write the VVF equation as F(t) = (x + at)i + (y + bt)j + (z + ct)k 0 0 0 Properties Given the VVFs F(t) and G(t) and the RVFs f and g, (F ▯ G)(t) = F(t) ▯ G(t) (F ▯ G)(t) = F(t) ▯ G(t) (F ▯ G)(t) = F(t) ▯ G(t) 1 (fF)(t) = f(t)F(t) (F ▯ g)(t) = F(g(t)) Example 1 Given F(t) = 3ti + 4j + 7t k and G(t) = 2tj + k ▯nd (F + G)(t). Solution We simply add the two VVFs component-wise in order to ascertain the result. (F + G)(t) = F(t) + G(t) = (3t + 0)i + (4 + 2t)j + (7t + 1)k 2 3ti + (4 + 2t)j + (7t + 1)k Example 2 Given H(t) = 2j + tk and J(t) = 3ti + j ▯nd (H ▯ J)(t). Solution We simply ▯nd the dot product of the two VVFs. (H ▯ J)(t) = H(t) ▯ J(t) = (0;2;t) ▯ (3t;1;0) (H ▯ J)(t) = (0)(3t) + (2)(1) + (t)(0) (H ▯ J)(t) = 2 2 Limits of Vector-Valued Functions Now we will look at how limits apply to VVFs. The following limit describes the velocity of a function at a point,0t lim F(t) ▯ F(t 0 = L t!t0 t ▯ 0 2 where L is some vector. If F(t) = (a(t);b(t);c(t)) then the velocity at t is (a(t) ▯ a(t 0;b(t) ▯ b(t )0c(t) ▯ c(t ))0 0 0 0 lim = (a (t 0;b (t);c (t 0) t!t 0 t ▯ t0 Properties The following are some properties of limits of VVFs. lim(F(t) ▯ G(t)) = lim F(t) ▯ lim G(t) t!t0 t!t0 t!t0 lim(F(t) ▯ G(t)) = lim F(t) ▯ lim G(t) t!t0 t!t0 t!t 0 lim(F(t) ▯ G(t)) = lim F(t) ▯ lim G(t) t!t0 t!t0 t!t0 lim(F(t) ▯ G(t)) = lim F(t) ▯ lim G(t) t!t0 t!t0 t!t0 t!t f(F(t)) = lit!t(t) limt!tt) 0 0 0 Example 3 Find lim t!▯ (tanti + 2tj ▯ 5k). Solution We ▯nd our answer by ▯nding the limit of each component lim(tanti + 2tj ▯ 5k) t!▯ lim(tant)i + lim(2t)k + lim(5)k t!▯ t!▯ t!▯ using some substitution we get 0i + 2▯j + 5k 2▯j + 5k 3 Example 4 2 2 Find lim t!2 (t ▯t▯2i + t ▯4t+4). t▯2 t▯2 Solution First we will ▯nd the limit of each component 2 2 ! lim t ▯ t ▯ 2 i + t ▯ 4t + 4 j t!2 t ▯ 2 t ▯ 2 ! ! t ▯ t ▯ 2 t ▯ 4t + 4 lim i + lim j t!2 t ▯ 2 t!2 t ▯ 2 and we will have to do some factoring in order to ▯nd the limits ! ! lim (t ▯ 2)(t + 1) i + lim (t ▯ 2)(t ▯ 2) j t!2 t ▯ 2 t!2 t ▯ 2 lim(t + 1)i + lim(t ▯ 2)j t!2 t!2 using some substitution we get 3i + 0j 3i 3 Derivatives and Integrals of VVFs The derivative of F(t) at t is0given formally by F(t) ▯ F(t )0 F (t0) = lim t!t0 t ▯ t0 If F(t) = f 1t)i + f 2t)j + f3(t)k, then F (t ) = f (t )i + f (t)j + f (t)k 0 1 0 2 3 4 Example 5 0 2 t Find F (5) if F(t) = 3t i + e j + 2k. Solution 0 F (t) can be determined by ▯nding the derivative of each component. F(t) = 3t i + e j + 2k F (t) = 6ti + e j and then we substitute 5 in for t. F (5) = 6(5)i + e j 30i + e j The integral of F(t) for an interval [a;b] is given by ! ! ! Z b Z b Z b Z b F(t)dt = f1(t)dt i + f 2t)dt j + f3(t)dt k a a a a Example 7 R2▯ 2 Find 0 F(t)dt if F(t) = t i + j + costk. Solution R2▯ 0 F(t)dt can be determined by ▯nding the integral of each component. F(t) = t i + j + costk Z b t3 F(t)dt = i + tj + sintk a 3 and substituting 0 and 2▯ for a and b, we get 3 (2▯) i + (2▯)j + sin2▯k ▯ (0i + 0j + 0k) 3 8▯ 3 + 2▯j 3 5 Properties Following are some properties of derivatives of VVFs. 0 0 0 (F ▯ G) (t) = F (t) ▯ G (t) 0 0 0 (F ▯ G) (t) = F (t) ▯ G(t) + F(t) ▯ G (t) (F ▯ G) (t) = F (t) ▯ G(t) + F(t) ▯ G (t) 0 0 0 0 (fF) (t) = f (t)F(t) + f(t)F (t) Example 6 Find ((t;0;5t ) ▯ (t ;e ;0)) . t 0 Solution Here, we just ▯nd the derivative of each VVF and subtract the two VVFs component-wise. 3 2 t 0 ((t;0;5t ) ▯ (t ;e ;0)) 2 t (1;0;15t ) ▯ (2t;e ;0) (1 ▯ 2t;0 ▯ e ;15t ▯ 0) 2 t 2 (1 ▯ 2t;▯e ;15t ) 6

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