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A telephone line hangs between two poles at 12m apart in the shape of a catenary y=50cos(x/50)-45, where x and y are measured in meters. Find the slope of this curve where it meets the right pole.
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In other words find the slope(dy/dx) when x=12 Differentiate the function y derivative of constant is zero, so the 45 just goes away for first term use chain rule: \[d/dx \cos u = -\sin u\] \[d/dx \frac{x}{50} = \frac{1}{50}\] \[\rightarrow d/dx 50\cos(x/50) = 50*\frac{1}{50}*-\sin(x/50) = -\sin(x/50)\] Now substitute in x=12 \[dy/dx = -\sin(12/50)= -0.2377\]
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