Stewart Calculus Answer Key

[mage lang="en|es|en" source="flickr"]stewart calculus answer key[/mage]
need a calculus genius! or has answer key to stewart calculus 5th edition?

does anybody have the complete solutions guide to stewart’s calculus 5th edition? i need help with #68 on page 715

or can you help me solve the problem?

show that the curves intersect at right angles.

r = a sin θ , r = a cos θ

thanks in advance :)

The point of intersection of the curves is given by
a sin θ = a cos θ
sin θ = cos θ
θ = pi/4

For the first curve,
r = a sinθ
x = r cosθ = a sinθ cosθ
dx/dθ = a [sinθ (-sinθ) + cosθ.cosθ]
dx/dθ = a(cos^2 θ – sin^2 θ) = a cos(2θ)
y = r sinθ = a sinθ sinθ = a sin^2 θ
dy/dθ = 2a sinθ cosθ = a sin(2θ)

dy/dx = dy/dθ / dx/dθ = [a sin(2θ)] / [a cos(2θ)]
dy/dx = tan(2θ)

At the point of intersection,
dy/dx = tan(2pi/4) = tan(pi/2) = infinity

Therefore the tangent to the first curve is vertical at the point of intersection.
_____________________________.

For the second curve,
r = a cosθ
x = r cosθ = a cosθ cosθ = a cos^2 θ
dx/dθ = a 2 cosθ (-sinθ)
dx/dθ = -a sin(2θ)

y = r sinθ = a cosθ sinθ
dy/dθ = a [cosθ cosθ + sinθ (-sinθ)]
dy/dθ = a[cos^2 θ - sin^2 θ]
dy/dθ = a cos(2θ)

dy/dx = dy/dθ / dx/dθ = [a cos(2θ)]/[-a sin(2θ)]
dy/dx = – cot(2θ)

At the point of intersection,
dy/dx = -cot(2pi/4) = -cot(pi/2) = -0 = 0

Therefore the tangent to the second curve is horizontal at the point of intersection.

Since the first curve is vertical and the second curve is horizontal at the point of intersection, therefore the curves intersect at right angles.

_________________.

[email protected]: Jean Hanff Korelitz

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