Using Stoke’s Theorem In Exercises 7-16, use Stoke’s Theorem to evaluate
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Calculus
- Using the Midpoint Formula Use the result of Exercise 53 to find the endpoint (x2,y2) of each line segment with the given endpoint (x1,y1) and midpoint (xm,ym). ax1,y1=1,2xm,ym=4,1bx1,y1=5,11xm,ym=2,4arrow_forwardLet f1(x)=3x and f2(x)=|x|. Graph both functions on the interval 2x2. Show that these functions are linearly dependent in the vector space C[0,1], but linearly independent in C[1,1].arrow_forwardSketch the space curve. Vector-Valued Function r(t) = -ti + 5tj + 2tk 5 5 2 5 Find its length over the given interval. Interval [0, 1] X X Z 2 5 2 X 5 5 V X ||arrow_forward
- Engineer math Determine the Partial Derivatives of the following Vectors with respect to x, y, and zarrow_forwardConsider the function θ : {0,1}×N→Z defined as θ(a,b) =a−2ab+b. Is θ injective? It is surjective? Bijective? Explain. Assume N to be the set of natural numbers and Z to be the set of integers.arrow_forwardCalculus Questionarrow_forward
- Exercise 2: Let f: X → Y and g: Y → Z be two bijective functions. Show that (gof)-1 = ƒ-1 o g¯1. Remark: This is frequently referred to as “shoes and socks" or something similar. Let f be the action of putting on socks, and g the action of putting on shoes. Then in order to get properly dressed, one ususally does go f: you put your socks on first, and then put your shoes on. However, at the end of the day, one does the opposite to undo this: one takes off their shoes first, and then their socks. Thus, (go f)-1 = f-1og¬1. Thus, the result makes sense. Note this is false for injective functions for a trivial reason, that the functions may not be possible to compose them.arrow_forwardProve the properties of the vector functions F and G and the scalar function f.arrow_forwardState whether this is true or false. If true, give an explanation. If false, give a counterexample.arrow_forward
- Consider the following function. T: R2 → R?, T(x, y) = (x, 5) Find the following images for vectors u = (u,, uz) and v = (v,, v,) in R2 and the scalar c. (Give all answers in terms of u,, u,, V, V2, and c.) T(u) = T(v) = T(u) + T(v) = T(u + v) = cT(u) = T(cu) = Determine whether the function is a linear transformation. O linear transformation O not a linear transformationarrow_forwardFind r(t) ∙ u(t). Is the result a vector-valued function? Explain. r(t) = (3t − 1)i + (1/4)t3 j + 4k, u(t) = t 2i − 8j + t 3karrow_forward|| Express the vector v = #+ [G] el y -[3] and [3] y= as a linear combination of = =arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:CengageElementary Linear Algebra (MindTap Course List)AlgebraISBN:9781305658004Author:Ron LarsonPublisher:Cengage Learning