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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.
What is derivative of cot t dt?
d/dt cot (t) dt = - cosec2(t)
What is the derivative of x equals cot2 divided by 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
What is hyperbolic functions?
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
Who is the sec of dot?
Dr. francisco T. Duque
What is 105-t?
It is an expression. There is nothing much you can do with an expression by itself.
What is t increased by 10?
The expression is: t+10
The velocity of a baseball 4 seconds after it is thrown vertically upward with a speed of 32.1 meter per sec is?
V = V0 + a t V0 = + 32.1 m/sec a = - 9.78 m/sec2 t = 4 sec V = (32.1) + 4 (-9.78) = (32.1) - (39.12) = - 7.02 m/sec (7.02 m/sec downward)
What are given in the problem?
f=3Hz T=0.33 sec. V=18m/s T=?
What is s fewer than t?
It is the expression t - s
What is an algebric expression for 47 less then t?
47< t
What is the algebraic expression for one-fifth of t?
t/5
What is t increased by 5 in algebraic expression?
t + 5
