Calculus Problem. Dealing with a limit of some trig functions.?
ø = Theta (because I do't have a theta on my keyboard) ^ = to the power of... ø->0 = as ø approaches 0 lim = the limit of... lim {(ø^2) (cosø)} / {(sinø)(tanø)} ø->0 If you let ø go to zero, sin(0) is zero and tan(0) is also zero. So you get a 0 in the denomonator and that is undefined. In order to get the limit, you have to maniplate the equation so that when you let ø go to 0, you don't get zero in the denominator. The question is: "Find the following limit"
Help with my pre calculus homework please, trigonometric functions?
I'm having problems to solve this problem please help: Instruction: Find the exact value of the trigonometric function at a given real number cot 8π/3= I know cot(t)=cos(t)/sin(t)=1/tan(t)=1/tan(t)=x... so I thought the answer was : -1/2 / √3/2 but it wasn't
Calculus: derivatives of inverse trigonometric functions?
we are aksed to find the first derivative of: Y = Cos^-1 (e^-T) (which reads as the inverse cosine of e to the power of negative T.) For my answer, I got: e^(-T) / sqrt((1 - e^(-2T))) Which read as e to the power of negative T, divided by the square root of 1 minus e to the power of negative 2T. I think I did this problem correctly, but I just wanted to make sure it's correct.
Calculus Help! Six trigonometric functions?
I need help with my homework... The problem is: tan x = root3/3 and cos x = -root3/2 I have to find the other four values. Cotangent and secant are easy. You just flip the fraction (and simplify) But I'm having trouble with sin. I know that tan= sin/cos... But if that was the case, wouldn't cos=3? it doesn't, because the problem says it's -root3/2... I remember back when we first learned this stuff, you had to find r= root x^2 + y^2 or something like that, but in this case, tangent (which is y/x) has a root in it, which messes me up... I know this is basic trig, but I'm having a brain block for some reason. Could someone please walk through the process? Thanks... first most helpful gets 10 points...
CALCULUS: Solve the problem by making use of inverse trigonometric functions?
Assuming that the leg is altitude of the isosceles triangle with reference to unequal side as base, we have base, B = 2*10*tan (θ/2), where "θ" is the apex angle having base B. dB/dt = (dB/dθ)*(dθ/dt) = 20*{sec^2 (θ/2)}*(1/2)*dθ/dt = 10*{sec^2 (θ/2)}*dθ/dt= -4 we have dθ/dt = -[4/{sec^2 (θ/2)] ---------------------- 1 when B is 16, tan (θ/2) = (16/2)/10 = 0.8 sec^2 (θ/2) = 1+ 0.8^2 = 1.64 or, the required (dθ/dt), when (B = 16 cm ) = = -[4/1.64] = -2.44 rad/s
CALCULUS: Solve the problem by making use of inverse trigonometric functions?
Let O be the searchlight on the ground and A the point on the ground where the object lands, so that OA=200. Let P be the position of the object so that AP is vertical, and OP the ray of light. You want the rate of change of the angle between OP and the vertical which = rate of change of angle OPA. =tan^-1(200/y) where AP=y d/dt(tan^-1(200/y))= [1/(1+ 200^2/y^2](-200/y^2)*dy/dt and when y=100 this =-0.004dy/dt. You now want some mechanics. The acceleration due to gravity is 32 ft/sec^2 and the speed (v ft/sec)with which the object drops a distance s ft from rest is given by v^2=64s. Here s=500-100=400 so v=160 . dy/dt =-160 (since y is decreasing) so -0.004dy/dt=0.64 rad/sec
Why are hyperbolic functions included in calculus instead of high school mathematics?
Things in math education tend to get introduced as they are needed, and not before.In high school, you study triangles, and the circular trig functions are useful for studying triangles — hence the definition of [math]\sin, \cos, \tan[/math] in relation to the sides of right triangles, not in relation to a unit circle. You might not hear of secant or cosecant, although you’ll probably hear of the cotangent. Pretty much all the work you do with trig functions uses degrees, and not radians.When you get to calculus, more aspects of trig functions are important, and you start working with them in relation to circles, not triangles. The entire notion of what sine, cosine, etc mean basically change.Hyperbolic functions are typically defined in terms of exponential functions — [math]\sinh x = \frac{e^x-e^{-x}}{2}, \cosh x = \frac{e^x+e^{-x}}{2}, \tanh = \frac{e^x-e^{-x}}{x^x+e^{-x}}[/math], and so on. The relationship between the functions and hyperbolae is not as clear as the relationship between the circular forms and circles.The first problem where hyperbolic functions are likely to show up in calculus is in the problem of a hanging rope or chain: given a chain of a known length suspended by the ends, what shape is the curve that it takes? The answer is a catenary, a hyperbolic cosine curve. Its a problem that can’t be solved without calculus, and in fact was an early calculus challenge problem.There’s no good reason to introduce hyperbolic functions before calculus; there’s no good problem that they solve before working with calculus, before working with exponential functions. In high school mathematics, the hyperbolic functions would be, at most, a curiosity, and example of weird functions that mathematicians can come up with with no real motivation.
CALCULUS: Solve the problems by making use of inverse trigonometric functions?
The distance from the light to the point where the ray hits the coast is r = √(0.5^2 + 1^2) by Pythagoras' theorem. The frequency of the light is 2 rpm --> 1 turn is 0.5 min. The tangential velocity at the distance of the radius r is 2pir/T v = 2pi*√(1.25)/0.5 mi/min = 4pi*√1.25 mi/min. The angle at the lighthouse between the perpendicular line to the coast and the point where the ray hits the coast, is tan^-1 (1/0.5) = 63.43°. And you can easily show, that the tangential velocity vector, when the ray hits the coast, also intersects with the cost line at an angle of 63.43°. The tangential magnitude of the velocity vector is 4pi*√1.25 mi/min at an angle of 63.43° with the coast line as we have seen. Now you have to find the component of this vector in direction of the coast line. For this you use the cosine function: cos(63.43°) = v/x where v is the tangential velocity, and x is the velocity along the coast line. --> x = v/cos(63.43°) = (4√(1.25)/cos(63.43))*pi x = 10.0 pi Regards
CALCULUS: Solve the following problems by making use of trigonometric functions...?
find the length of the shortest ladder which will reach from the ground level to a high vertical wall if it must clear an 8ft vertical fence which is 27ft from the wall. the answer should be 13 sqrt(13) show the solution and provide illustration please. 5 start for the best answer