d/dt cot (t) dt = - cosec2(t)
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
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.
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.
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
d/dt cot (t) dt = - cosec2(t)
Assuming you want dx/dt and that the equation is x = cot(2) / t (i.e. cot(2) is a constant) we can use the power rule. First, we rewrite it: cot(2)/t = cot(2) * t-1 thus, by the power rule: dx/dt = (-1) cot(2) * t-1 -1 = - cot(2) * t-2= = -cot(2)/t2
There is sufficient information to determine Jack's speed at t equals 10 seconds.
It is an expression. There is nothing much you can do with an expression by itself.
The expression is: t+10
Cot curve is concerned with the measurement of the degree of reannealing of DNA strands. It is a curve drawn with X-axis having DNA concentration unit multiplied by time. Since the initial concentration is considered represented as Co and when multiplied with time t, it becomes "Cot" and the graph is known as Cot curve. The graph is drawn against %reanealled versus Cot.
Dr. francisco T. Duque
The hyperbolic functions are related to a hyperbola is the same way the the circular functions are related to a circle. So, while the points with coordinates [cos(t), sin(t)] generate the unit circle, their hyperbolic counterparts, [cosh(t) , sinh(t)] generate the right half of the equilateral hyperbola. Other circular functions (tan, sec, cosec and cot) also have their hyperbolic counterparts, as do the inverse functions. An alternative, equivalent pair of definitions is: cosh(x) = (ex + e-x)/2 and sinh(x) = (ex - e-x)/2
t + 5
It is the expression t - s