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Find the linear approximation to f(x) = cos(2x) at x = π/6. Use the linear approximation to approximate the value of cos(1/2). Please enter your answer in decimal format with three significant digits after the decimal point. . A person 2 m tall walks towards a lamppost on level ground at a rate of 0.6m/sec. The lamp on the post is 6m high. At which rate the length of the person's shadow decreasing when the person is 3m from the post?

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Final answer:

The linear approximation of cos(2x) at x = π/6 is used to approximate the value of cos(1/2). To find the rate at which the length of a person's shadow is decreasing when they are 3m from a lamppost, we use related rates involving similar triangles.

Step-by-step explanation:

Linear Approximation and Rate of Change

To find the linear approximation of f(x) = cos(2x) at x = π/6, we use the formula L(x) = f(a) + f'(a)(x - a), where L(x) is the linear approximation, a is the point of tangency, x is the point we want to approximate, and f'(a) is the derivative of f at a. First, we calculate f(π/6) and the derivative f'(π/6). Since cos(2π/6) = √3/2 and the derivative of cos(2x) is -2sin(2x), we have f'(π/6) = -2sin(π/3) which equals -√3. Thus, the linear approximation near x = π/6 is L(x) = √3/2 - √3(x - π/6). To approximate cos(1/2), we simply substitute x = 1/2 into the linear approximation.



For the related rates problem, if a person walks towards a lamppost with a speed of 0.6 m/s, and the person is 3m from the post, we can use similar triangles to find the rate at which the shadow's length is changing. Denoting the person's height by h_p, the lamppost's height by h_l, the person's distance to the post by d, and the shadow's length by s, we have the ratio h_p / (s + d) = h_l / d. Differentiating both sides with respect to time t, we get an equation that allows us to solve for ds/dt, the rate at which the shadow's length is decreasing.

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