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# Calculus Help

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Find the absolute maximum and absolute minimum values of f in the given interval.

f(t)=t+cot(t/2) [pi/4, 7pi/4]

Please show how you got your answer by finding the critical numbers by finding the derivative  of the equation first and then setting the first  derivative equal to zero.  Then find the points where t=pu/4 and t=7pi/4.

gibsonj338  Jun 10, 2018
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Hi Gibson338 :)

Find the absolute maximum and absolute minimum values of f in the given interval.

f(t)=t+cot(t/2) [pi/4, 7pi/4]

pi/4 = 0.7853981633974483

7pi/4 = 5.4977871437821382

end points

f(pi/4)=pi/4+cot(pi/8)          =   pi/4+cot(pi/8) = 3.20

f(7pi/4)= 7pi/4+cot(7pi/8)   =    7pi/4+cot(7pi/8) = 3.08

end points function values are approx 3.2  and 3.08

$$\text{Firstly cot(t/2) must be defined}\\ so\;\; sin(t/2) \ne0\\ so \;\;t/2\ne n\pi \qquad where\;\;n\in Z\\ so \;\;t\ne 2n\pi \qquad where\;\;n\in Z$$

$$f(t)=t+cot(t/2)\\ let\;\;y=f(t)\\ y=t+\frac{cos(t/2)}{sin(t/2)}\\ let\\ u=cos(t/2) \qquad \\ \frac{du}{dt}=0.5*-sin(t/2)\\ \frac{du}{dt}=-0.5sin(t/2)\\~\\ v=sin(t/2)\\ \frac{dv}{dt}=0.5*cos(t/2)\\~\\ \frac{dy}{dt}=1+\frac{vu'-uv'}{v^2}\\ \frac{dy}{dt}=1+\frac{sin(t/2)[-0.5sin(t/2)]-cos(t/2)0.5(cos(t/2)}{sin(t/2)^2}\\ \frac{dy}{dt}=1+\frac{-0.5sin^2(t/2)-0.5(cos^2(t/2)}{sin(t/2)^2}\\ \frac{dy}{dt}=1+\frac{-0.5}{sin(t/2)^2}\\$$

Find stat points    that is when dy/dt = 0

$$0=1+\frac{-0.5}{sin(t/2)^2}\\ 1=\frac{0.5}{sin(t/2)^2}\\ sin(t/2)^2=0.5\\ sin(t/2)=\pm\sqrt{0.5}\\ \frac{t}{2}= n\pi \pm \frac{\pi}{4}\\ t= 2\pi n\pm \frac{\pi}{2}\\$$

but

$$\frac{\pi}{4} \le t \le \frac{7\pi}{4}\\ t=\frac{\pi}{2}\;\;or \;\;\frac{3\pi}{2}\\ \text{So the stationary points are }\\ (\frac{\pi}{2},2.571),\quad (\frac{3\pi}{2},3.712)$$

$$end\; points\; ( \frac{\pi}{2}, 3.20) \;\;and \;\; ( \frac{7\pi}{4}, 3.08)$$

So the absolute minimum is approx 2.571 when t=pi/2

and the absolute maximum is approx 3.712 when t=3pi/2

Here is the graph:

Melody  Jun 10, 2018

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