A calculus problem by Rocco Dalto

Calculus Level pending

The y y axis and the line x = c x = c are the banks of a river whose current has uniform speed a \bf a in the negative y y direction. A boat enters the river at the point ( c , 0 ) (c,0) and heads directly toward the origin with speed b \bf b relative to the water.

What are the conditions on a \bf a and b \bf b that will allow the boat to reach the opposite bank? Where will it land?

(1) a < b \bf a < b and the boat will land at the origin ( 0 , 0 ) (0,0) .

(2) a = b \bf a = b and the boat will land at ( 0 , − c 2 ) (0,-\dfrac{c}{2}) .

(3) a > b \bf a > b and the boat will land at ( 0 , − c ) (0, -c)

(4) a > b \bf a > b and the boat will never land.

(5) a < b \bf a < b and the boat will never land.

(6) a = b \bf a = b and the boat will never land.

Only (2),(4), and (5) are true Only (1) ,(2), and (4) are true Only (1) ,(2), and (3) are true. Only (1),(4), and (6) are true Only (2),(3), and (5) are true Only (3),(5), and (6) are true.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Dec 12, 2016

The components of the velocity are:

d x d t = − b c o s θ \dfrac{dx}{dt} = -b cos\theta d y d t = − a + b s i n θ \dfrac{dy}{dt} = -a + b sin\theta

Let k = a b ⟹ d y d x = k s e c θ − t a n θ x ⟹ k = \dfrac{a}{b} \implies \dfrac{dy}{dx} = \dfrac{k sec\theta - tan\theta}{x} \implies

d y d x = k x 2 + y 2 + y x , \dfrac{dy}{dx} = \dfrac{k \sqrt{x^2 + y^2} + y}{x}, where s e c θ = x 2 + y 2 x sec\theta = \dfrac{\sqrt{x^2 + y^2}}{x} and t a n θ = − y x . tan\theta = \dfrac{-y}{x}.

The differential above is homogeneous of degree zero.

Let z = y x ⟹ y = x z ⟹ d y = x d z + z d x ⟹ x d z + z d x = k ( 1 + z 2 + z ) d x ⟹ z = \dfrac{y}{x} \implies y = xz \implies dy = x dz + z dx \implies x dz + z dx = k (\sqrt{1 + z^2} + z) dx \implies

x d z = k 1 + z 2 ⟹ 1 k ∫ d z 1 + z 2 = ∫ d x x x dz = k \sqrt{1 + z^2} \implies \dfrac{1}{k} \int \dfrac{dz}{\sqrt{1 + z^2}} = \int\ \dfrac{dx}{x}

Let z = t a n λ ⟹ d z = s e c 2 ( λ ) d λ ⟹ 1 k ln ⁡ ( s e c λ + t a n λ ) + j = z = tan\lambda \implies dz = sec^2(\lambda) \: d\lambda \implies \dfrac{1}{k} \ln(sec\lambda + tan\lambda) + j =

1 k ln ⁡ ( z 2 + 1 + z ) + j = ln ⁡ ( x ) \dfrac{1}{k} \ln(\sqrt{z^2 + 1} + z) + j = \ln(x)

z = y x ⟹ 1 k ln ⁡ ( x 2 + y 2 + y x ) + j = ln ⁡ ( x ) z = \dfrac{y}{x} \implies \dfrac{1}{k} \ln(\dfrac{\sqrt{x^2 + y^2}+ y}{x}) + j = \ln(x)

y ( c ) = 0 ⟹ j = ln ⁡ ( c ) ⟹ ln ⁡ ( x 2 + y 2 + y x ) = ln ⁡ ( x k c k ) ⟹ ln ⁡ ( c k ∗ ( x 2 + y 2 + y ) x k + 1 ) = 0 y(c) = 0 \implies j = \ln(c) \implies \ln(\dfrac{\sqrt{x^2 + y^2} + y}{x}) = \ln(\dfrac{x^k}{c^k}) \implies \ln(\dfrac{c^k * (\sqrt{x^2 + y^2} + y)}{x^{k + 1}}) = 0

⟹ c k ∗ ( x 2 + y 2 + y ) = x k + 1 \implies c^k * (\sqrt{x^2 + y^2} + y) = x^{k + 1}

Solving for y y we have:

( x 2 + y 2 + y ) 2 = ( x k + 1 c k ) 2 ⟹ (\sqrt{x^2 + y^2} + y)^2 = (\dfrac{x^{k + 1}}{c^k})^2 \implies

2 x 2 + y 2 ∗ y = ( ( x k + 1 c k ) 2 − x 2 ) − 2 y 2 ⟹ 2 \sqrt{x^2 + y^2} * y = ((\dfrac{x^{k + 1}}{c^k})^2 - x^2) - 2 y^2 \implies

4 x 2 y 2 + 4 y 4 = ( ( x k + 1 c k ) 2 − x 2 ) 2 − 4 ∗ ( ( x k + 1 c k ) 2 − x 2 ) ∗ y 2 + 4 y 4 ⟹ 4 x^2 y^2 + 4 y^4 = ((\dfrac{x^{k + 1}}{c^k})^2 - x^2)^2 - 4 * ((\dfrac{x^{k + 1}}{c^k})^2 - x^2) * y^2 + 4 y^4 \implies

0 = ( ( x k + 1 c k ) 2 − x 2 ) 2 − 4 ∗ ( ( x k + 1 c k ) 2 y 2 ) 0 = ((\dfrac{x^{k + 1}}{c^k})^2 - x^2)^2 - 4 * ((\dfrac{x^{k + 1}}{c^k})^2 y^2)

Choosing the positive value for y we obtain:

y = 1 2 ( c k x k + 1 ) ∗ ( x 2 ( k + 1 ) c 2 k − x 2 ) ⟹ y = \dfrac{1}{2} (\dfrac{c^k}{x^{k + 1}}) * (\dfrac{x^{2(k + 1)}}{c^{2k}} - x^2 ) \implies

y ( x ) = 1 2 ∗ ( x k + 1 c k − c k x k − 1 ) y(x) = \dfrac{1}{2} * (\dfrac{x^{k + 1}}{c^k} - \dfrac{c^k}{x^{k - 1}})

For a > b ( k > 1 ) , l i m x → 0 y ( x ) = − ∞ ⟹ a > b (k > 1), lim_{x \rightarrow 0} y(x) = -\infty \implies boat will never land.

For a = b ( k = 1 ) , l i m x → 0 y ( x ) = − c 2 ⟹ a = b (k = 1), lim_{x \rightarrow 0} y(x) = -\dfrac{c}{2} \implies boat will land at ( 0 , − c 2 ) . (0,-\dfrac{c}{2}).

For a < b ( k < 1 ) , l i m x → 0 y ( x ) = 0 ⟹ a < b (k < 1), lim_{x \rightarrow 0} y(x) = 0 \implies boat will land at the origin ( 0 , 0 ) . (0,0).

∴ \therefore Only (1), (2), and (4) are true.

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