Question

The Eiffel Tower Property Let R be the region between the curves y = e^(-c·x) and y = -e^(-c·x) on the interval [a, ∞), where a ≥ 0 and c > 0.
The center of mass of R is located at (x̄, 0), where x̄ = [∫(a to ∞) x·e^(-c·x) dx] / [∫(a to ∞) e^(-c·x) dx]
(The profile of the Eiffel Tower is modeled by these two exponential curves; see the Guided Project ""The exponential Eiffel Tower"")
b. With a = 0 and c = 2, find the equations of the lines tangent to both curves at x = 0

Accepted Answer

Identify the two curves given: the upper curve is $$y = e$$^{-2x}$$ and the lower curve is y$$ = $$-e^{-2x}$$, since $$a = 0$$ and $$c = 2. $$Find the derivative of the upper curve $$y = e$$^{-2x}$$ with respect to x$ to determine the slope of the tangent line at x = 0$. Use the chain rule: $$\frac{dy}{dx} = $$-2e^{-2x}. $$Evaluate the slope of the tangent line at $$x = 0 by $$substituting $$x = 0$$ into the derivative: $$m = -2e$$^{0}$$ = -2. $$Find the point of tangency on the upper curve at $$x = 0$$: $$y = e$$^{0}$$ = 1. So $$the point is $$(0, 1). $$Use the point-slope form of a line $$y - y_1$ = m(x - x_1$)$$ with the point $$(0, 1)$$ and slope $$-2 to $$write the equation of the tangent line to the upper curve. Repeat the process for the lower curve $$y = -e$$^{-2x}$$, noting that its derivative is $$\frac{dy}{dx} = $$2e^{-2x}$$ and the point at $$x=0 is$$ $$(0, -1).$$

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