Areas of Simple Closed Curves In Exercises 81-86, use a computer algebra system and the result of Exercise 77 to match the closed curse with its area. (These exercises were based on “The Surveyor’s Area Formula'’ by Bart Braden, College Mathematics Journal, September 1986, pp. 335-337, by permission of the author.)
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Hourglass:
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CALCULUS
- 1. Find the area of the region enclosed between the curves y = x and y = x. Sketch the region.arrow_forwardfor the given rectangular coordinates, find two sets of polar coordinates for which 0≤θ<2π, one with r>0 and the other with r<0. (-2sqrt(3),9)arrow_forwardI circled the correct answer, could you show me how to do it using divergence and polar coordinatesarrow_forward
- The correct answer is D Could you explain and show the steps pleasearrow_forwardTaylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52arrow_forwardCould you explain this using the formula I attached and polar coorindatesarrow_forward
- Let g(z) = z-i z+i' (a) Evaluate g(i) and g(1). (b) Evaluate the limits lim g(z), and lim g(z). 2-12 (c) Find the image of the real axis under g. (d) Find the image of the upper half plane {z: Iz > 0} under the function g.arrow_forwardk (i) Evaluate k=7 k=0 [Hint: geometric series + De Moivre] (ii) Find an upper bound for the expression 1 +2x+2 where z lies on the circle || z|| = R with R > 10. [Hint: Use Cauchy-Schwarz]arrow_forward21. Determine for which values of m the function (x) = x™ is a solution to the given equation. a. 3x2 d²y dx² b. x2 d²y +11x dy - 3y = 0 dx dy dx2 x dx 5y = 0arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage