Question

1. Suppose that $C$ is the top-half of the circle $(x - a)^2 + y^2 = a^2$, ($a > 0$) oriented in the counterclockwise direction. Let $F$ be a force field that has constant magnitude $c$ and always directed from $(x, y)$ toward the origin. The work done by the force $F$ on a particle that moves along $C$ is $\int_C \mathbf{F} \cdot \mathbf{T} \, ds = 2ac$. Explain how do we obtain this formula. (6 marks CO2 (3%))

          1. Suppose that $C$ is the top-half of the circle $(x - a)^2 + y^2 = a^2$, ($a > 0$) oriented in
the counterclockwise direction. Let $F$ be a force field that has constant magnitude $c$
and always directed from $(x, y)$ toward the origin. The work done by the force $F$ on a
particle that moves along $C$ is
$\int_C \mathbf{F} \cdot \mathbf{T} \, ds = 2ac$.
Explain how do we obtain this formula.
(6 marks CO2 (3%))

1. Suppose that C is the top-half of the circle (x - a)^2 + y^2 = a^2, (a > 0) oriented in
the counterclockwise direction. Let F be a force field that has constant magnitude c
and always directed from (x, y) toward the origin. The work done by the force F on a
particle that moves along C is
𝐅·𝐓 ds = 2ac.
Explain how do we obtain this formula.
(6 marks CO2 (3%))

Added by Daniela T.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Hemraj Kumawat

Step 1

First, let's find the equation of the curve C. The equation of a circle with center (a,0) and radius a is given by (x - a)^2 + y^2 = a^2. Since we are only interested in the top-half of the circle, we can rewrite the equation as y = sqrt(a^2 - (x - a)^2). Show more…

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Suppose that C is the top-half of the circle (x - a)^2 + y^2 = a^2, a > 0, oriented in the counterclockwise direction. Let F be a force field that has constant magnitude c and is always directed from (y) toward the origin. The work done by the force F on a particle that moves along C is Tds = 2ac. Explain how we obtain this formula. (6 marks CO2 (3%))