A curve is given by on and has a vertical tangent at with

MATHEMATICS · Single correct · AP · Calculus BC · Applications of Integration · Applications of Integration

Problem

A curve is given by

y=f(x)

[a,b]

and has a vertical tangent at

x=k

with

a<k<b

. Which of the following best explains why re-expressing the curve as

x=g(y)

can simplify the computation of its length? (a) Because

\dfrac{dy}{dx}=0

at a vertical tangent, the integrand

\sqrt{1+\bigl(\tfrac{dy}{dx}\bigr)^{2}}

simplifies to

1

, so the arc-length integral collapses to the length of the interval. (b) Using

x=g(y)

shortens the interval of integration, which always reduces the difficulty of evaluating the arc-length integral. (c) Near a vertical tangent,

\dfrac{dy}{dx}

becomes unbounded, whereas

\dfrac{dx}{dy}

remains finite, so re-expressing the curve keeps the integrand of the arc-length integral bounded. (d) The arc-length formula written in terms of

y=f(x)

is not valid whenever the curve possesses a vertical tangent, so one must switch to

x=g(y)