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What is derivative of cot t dt?
d/dt cot (t) dt = - cosec2(t)
A light is hung 15 feet above a straight horizontal path If a man 6 feet tall is walking away from the light at the rate of 5 feet per second how fast is his shadow lenghtening?
Suppose the distance from the light to the man at time t is x(t), and that the length of his shadow is s(t).By similar triangles, (x + s)/s = 15/6 = 5/2That is, x/s + 1 = 5/2 so that x/s = 3/2and therefore, s = 2x/3Then since dx/dt = 5 feet/sec, ds/dt = 2/3*5 ft/sec = 10/3 ft/sec or 3.33... (recurring) ft/sec
What is derivative of 9et?
Assume that the expression is: y = 9e^(t) Remember that the derivative of e^(t) with respect to t is e^(t). If we take the derivative of the function y, we have.. dy/dt = 9 d[e^(t)]/dt = 9e^(t) Note that I factor out the constant 9. If we keep the 9 in the brackets, then the solution doesn't make a difference.
A car braked with a constant deceleration of 16 ft per sec squared producing skid marks measuring 200ft before coming to a stop How fast was the car traveling when the brakes were first applied?
If you have taken physics, you should know the basic kinematic formula vf²=vi²+2ad 0=vi²-2*16*200 vi=80m/s Alternatively, if you haven't taken physics yet, you will have to derive an equivalent formula on your own. We know that the integral of acceleration is equal to velocity so integrate -16 with respect to time to get v=-16t + vi for some constant vi. We also know that the integral of velocity with respect to time is distance so d=-8t²+vi*t+c for a constant value c. Since the distance traveled by the car was 0 at t=0, the value of c is 0. We also know that there was a constant acceleration so a=(vf-vi)/t -16=-vi/t vi=16t Combining this with the preceding equation for distance gives us d=-8t²+vi*t 200=-8t²+16t² t=5s (reject the negative root) vi=16t vi=80m/s
Integration of tangent cubed of x?
Find I = ∫ tan³ x dx. The solution is: I = ½ tan² x - log cos x. * * * Here is how we can obtain this result: First, let t = tan x, s = sin x, and c = cos x; then, dI = t³ dx, ds = c dx, dc = -s dx, and dt = (1 + t²) dx; and, of course, t = s / c. By algebra, t³ = t(t² + 1) - t; thus, we have dI = t³ dx = t(t² + 1) dx - t dx = t dt - t dx. Now, d (t²) = 2t dt; thus, t dt = ½ d(t²). On the other hand, we have d log c = dc / c = -s dx / c = -t dx; thus, t dx = -d log c. Combining these results, we have dI = t dt - t dx = ½ d(t²) - d log c. This integrates readily, giving I = ½ t² - log c, which is the solution we sought. * * * We may check our result, by differentiating back: dt / dx = 1 + t²; and d(t²) / dt = 2t; thus, (d/dx)(t²) = 2t dt / dx = 2t (1 + t²). Also, we have d log c / dc = 1 / c; and dc / dx = -s; whence, (d/dx)(log c) = (dc / dx) / c = -s / c = -t. Then, dI / dx = ½ (d/dx)(t²) - (d/dx)(log c) = t (1 + t²) - t = t + t³ - t = t³, re-assuring us that we have integrated correctly.
