SOLUTION 16: $\ \$ A motorboat is resting at the origin, $(0, 0)$, and a skier tethered to the boat with a 20-foot long rope, is resting at the point $(20, 0)$. (See the diagram below.)

The boat then begins moving along the positive $y$-axis, pulling the skier along the unknown path $y=f(x)$. Use integration to find an equation for this path. We will assume that the 20-foot rope lies on a tangent line to the unknown curve $y=f(x)$ at the point of tangency at $(x,y)$ (The point $(x,y)$ represents the skier.). Consider the right triangle formed by the boat on the $y$-axis, the skier on the curve at the point $(x,y)$, and the point $(0,y)$. The base of this right triangle is $x$. The hypotenuse of this right triangle is 20. Let $L$ be the height of this right triangle. (See the diagrams below.)

By the Pythagorean Theorem we get that $$x^2+L^2 = 20^2 \ \ \longrightarrow \ \ \ L = \sqrt{ 400-x^2 }$$ Since the hypotenuse is on the tangent line to the unknown curve $y=f(x)$, it follows that the slope of the tangent line is $$f'(x) = slope = \displaystyle {\frac {rise}{run}} = \displaystyle {\frac {-L}{x}} = \displaystyle {\frac {-\sqrt{ 400-x^2 }}{x}}$$ so that the unknown curve is $$f(x) = \displaystyle { \int \frac {-\sqrt{ 400-x^2 }}{x} \ dx } = \displaystyle { - \int \frac {\sqrt{ 400-x^2 }}{x} \ dx }$$ together with the fact that $f(20)=0$ (position of skier before boat takes off). Begin the integration with the trig substitution $$x = 20 \sin \theta$$ so that $$dx = 20 \cos \theta \ d \theta$$ Substitute into the original problem, replacing all forms of x, getting $$\displaystyle { - \int \frac {\sqrt{ 400-x^2 }}{x} \ dx } = \displaystyle { - \int \frac {\sqrt{ 400-(20 \sin \theta)^2 }}{20 \sin \theta} \cdot 20 \cos \theta \ d \theta }$$ $$= \displaystyle { - \int \frac {\sqrt{ 400-400 \sin^2 \theta }}{ \sin \theta} \cdot \cos \theta \ d \theta }$$ $$= \displaystyle { - \int \frac {\sqrt{ 400(1- \sin^2 \theta) }}{ \sin \theta} \cdot \cos \theta \ d \theta }$$ $$= \displaystyle { - \int \frac {20 \sqrt{1- \sin^2 \theta }}{ \sin \theta} \cdot \cos \theta \ d \theta }$$ $$= \displaystyle { - 20 \int \frac { \sqrt{ \cos^2 \theta }}{ \sin \theta} \cdot \cos \theta \ d \theta }$$ $$= \displaystyle { - 20 \int \frac { \cos \theta }{ \sin \theta} \cdot \cos \theta \ d \theta }$$ $$= \displaystyle { - 20 \int \frac { \cos^2 \theta }{ \sin \theta} \ d \theta }$$ (Recall that $\cos^2 \theta + \sin^2 \theta = 1$ so that $\cos^2 \theta = 1 - \sin^2 \theta$.) $$= \displaystyle { - 20 \int \frac { 1 - \sin^2 \theta }{ \sin \theta} \ d \theta }$$ $$= \displaystyle { - 20 \int \Big( \frac { 1 }{ \sin \theta} - \frac { \sin^2 \theta }{ \sin \theta} \Big) \ d \theta }$$ $$= \displaystyle { - 20 \int ( \csc \theta - \sin \theta ) \ d \theta }$$ $$= \displaystyle { - 20 ( \ln \Big| \csc \theta - \cot \theta \Big| + \cos \theta ) + C }$$ $\Big($ We need to write our final answer in terms of $x$.

Since $x = 20 \sin \theta$ it follows that $$\sin \theta = \displaystyle{ x \over 20 } = \displaystyle{ opposite \over hypotenuse }$$ and from the Pythagorean Theorem that $$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(adjacent)^2 + (x)^2 = (20)^2 \ \ \longrightarrow \ \ \ adjacent = \sqrt{400-x^2} \ \ \longrightarrow$$ $$\cos \theta = \displaystyle{ adjacent \over hypotenuse }= \displaystyle{ \sqrt{400-x^2} \over 20 }$$ $$\cot \theta = \displaystyle{ adjacent \over opposite }= \displaystyle{ \sqrt{400-x^2} \over x }$$ and $$\csc \theta = \displaystyle{ hypotenuse \over opposite }= \displaystyle{ 20 \over x } . \Big)$$ $$= \displaystyle { - 20 ( \ln \Big| { 20 \over x } - { \sqrt{400-x^2} \over x } \Big| + { \sqrt{400-x^2} \over 20 } ) + C }$$ $$= \displaystyle { - 20 \ln \Big| { 20 - \sqrt{400-x^2} \over x } \Big| - \sqrt{400-x^2} + C }$$ i.e., $$f(x) = \displaystyle { - 20 \ln \Big| { 20 - \sqrt{400-x^2} \over x } \Big| - \sqrt{400-x^2} + C }$$ Since $f(20)=0$, we get that $$f(20) = \displaystyle { - 20 \ln \Big| { 20 - \sqrt{400-(20)^2} \over 20 } \Big| - \sqrt{400-(20)^2} + C } \ \ \longrightarrow$$ $$0 = \displaystyle { - 20 \ln \Big| { 20 - 0 \over 20 } \Big| - \sqrt{400-(20)^2} + C } \ \ \longrightarrow$$ $$0 = \displaystyle { - 20 \ln 1 - \sqrt{0} + C } \ \ \longrightarrow$$ $$0 = \displaystyle { - 20 (0) + C } \ \ \longrightarrow$$ $$C=0$$ Thus, the skier follows the curve given by $$f(x) = \displaystyle { - 20 \ln \Big| { 20 - \sqrt{400-x^2} \over x } \Big| - \sqrt{400-x^2} }$$

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