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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

(a) If C is the line segment connecting the point (x1, y1) to the point (x2, y2), show that

C x dyy dx = x1y2x2y1

(b) If the vertices of a polygon, in counterclockwise order, are (x1, y1), (x2, y2), , (xn, yn), show that the area of the polygon is

A = 1 2 [(x1, y2), (x2, y1) + (x2y3x3y2) + ⋯ + (xn−1 ynxn yn−1) + (xny1 x1yn)]

(c) Find the area of the pentagon with vertices (0, 0), (2, 1), (1, 3), (0, 2), and (−1, 1).

(a)

To determine

To show: For line segment connecting point (x1,y1) to point (x2,y2) , Cxdyydx=x1y2x2y1 .

Explanation

Given data:

Points are (x1,y1) and (x2,y2) .

Formula used:

Write the expression for line segment which starts at vector r0 and terminates at vector r1 .

r(t)=(1t)r0+tr1 (1)

Here,

t is parameter.

Write the parametric equations of line segment C, 0t1 with start point (x1,y1) and terminal point (x2,y2) by the use of equation (1).

x=(1t)x1+tx2 (2)

y=(1t)y1+ty2 (3)

Differentiate equation (2) with respect to t.

ddt(x)=ddt((1t)x1+tx2)dxdt=ddt((1t)x1)+ddt(tx2)dxdt=(01)x1+(1)x2{ddt(k)=0,ddt(t)=1}dxdt=x2x1

dx=(x2x1)dt

Differentiate equation (3) with respect to t.

ddt(y)=ddt((1t)y1+ty2)dydt=ddt((1t)y1)+ddt(ty2)dydt=(01)y1+(1)y2{ddt(k)=0,ddt(t)=1}dydt=y2y1

dy=(y2y1)dt

Find the value of Cxdyydx

(b)

To determine

To show: The area of polygon with vertices (x1,y1),(x2,y2),...,(xn,yn) as A=12[(x1y2x2y1)+(x2y3x3y2)+...+(xn1ynxnyn1)+(xny1x1yn)] .

(c)

To determine

To find: The area of pentagon with vertices (0,0) , (2,1) , (1,3) , (0,2) , and (1,1) .

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Chapter 16 Solutions

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