A steady wind blows a kite due west The kites height above g

A steady wind blows a kite due west. The kite\'s height above ground from horizontal position x = 0 to x = 100 ft is given by
y = 150-(1/40)(x-50)^2.
Find the distance traveled by the kite. (Round your answer to one decimal place.)

y= 150 - (1/40)(x-50)^2 dy/dx = -(1/40)(2)(x-50) = (50-x)/20 (dy/dx)^2 = (50-x)^2/400 1+(dy/dx)^2 = 1+(50-x)^2/400 = [400+(50-x)^2] / 400 sqrt[1+(dy/dx)^2] = 1/20 sqrt[(50-x)^2+20^2) Arc length = (1/20) ? sqrt((50-x)^2+20^2) dx , 0 <= x <= 80 Integrating using Simpson\'s Rule with n=30 subdivisions: Arc length = 122.7761 Note: You may try to integrate it by using the substitution 50-x = sinh(u) , where sinh(u) is the hyperbolic sine function.

$A steady wind blows a kite due west. The kite\'s height above ground from horizontal position x = 0 to x = 100 ft is given by y = 150-(1/40)(x-50)^2. Find the d$

A steady wind blows a kite due west The kites height above g

Solution

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