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Abdelrahman T.
Mathematics Contributor at Slader LLC
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Trigonometry
TutorMe
Question:

An object is propelled upward at an angle to the horizontal with an initial velocity of feet per second from the base of a plane that makes an angle of $$45^{o}$$ with the horizontal. See the illustration. If air resistance is ignored, the distance R that it travels up the inclined plane is given by the function $$R(\theta) =\dfrac{v_{o}^{2}\sqrt{2}}{16}\cos(\theta)(\sin(\theta)-\cos(\theta))$$ Show that $$R(\theta) =\dfrac{v_{o}^{2}\sqrt{2}}{32}[\sin(2\theta)-\cos(2\theta)-1]$$

Abdelrahman T.
Answer:

We have to prove two formulas are equivalent, thus we begin with the equation: $$R(\theta) =\dfrac{v_{o}^{2}\sqrt{2}}{16}\cos(\theta)(\sin(\theta)-\cos(\theta))$$ $$ \to \quad R(\theta) =\dfrac{v_{o}^{2}\sqrt{2}}{16}(\cos(\theta)\sin(\theta)-\cos^2(\theta))$$ $$ \to \quad R(\theta) =\dfrac{v_{o}^{2}\sqrt{2}}{16}(\frac{1}{2}\cdot 2\cos(\theta)\sin(\theta)-\frac{1}{2}\cdot 2\cos^2(\theta))$$ $$ \to \quad R(\theta) =\dfrac{v_{o}^{2}\sqrt{2}}{16}[\frac{1}{2}\cdot \sin(2\theta)-\frac{1}{2}\cdot 2\cos^{2}(\theta)] \qquad $$[Using: $$\sin(2\theta)=2\sin(\theta)\cos(\theta)$$] $$ \to \quad R(\theta) =\dfrac{v_{o}^{2}\sqrt{2}}{16}[\frac{1}{2}\cdot \sin(2\theta)-\frac{1}{2}\cdot (\cos(2\theta)+1)] \qquad $$[Using: $$\cos(2\theta)=2\cos^2(\theta)-1$$] $$ \to \quad R(\theta) =\dfrac{v_{o}^{2}\sqrt{2}}{16}[\frac{1}{2}\cdot \sin(2\theta)-\frac{1}{2}\cdot \cos(2\theta)-\frac{1}{2}] $$ $$ \to \quad R(\theta) =\dfrac{1}{2} \cdot \dfrac{v_{o}^{2}\sqrt{2}}{16}[\sin(2\theta)- \cos(2\theta)-1] $$ $$ \to \quad R(\theta) =\dfrac{v_{o}^{2}\sqrt{2}}{32}[\sin(2\theta)- \cos(2\theta)-1] $$ Thus, the two formulas are equal

Differential Equations
TutorMe
Question:

A mass weighing 64 pounds stretches a spring 0.32 foot. The mass is initially released from a point 8 inches above the equilibrium position with a downward velocity of 5 ft/s. (a) Find the equation of motion. (b) What are the amplitude and period of motion? (c) How many complete cycles will the mass have completed at the end of 3p seconds? (d) At what time does the mass pass through the equilibrium position heading downward for the second time? (e) At what time does the mass attain its extreme displacement on either side of the equilibrium position? (f ) What is the position of the mass at t = 3 s? (g) What is the instantaneous velocity at t = 3 s? (h) What is the acceleration at t = 3 s? (i) What is the instantaneous velocity at the times when the mass passes through the equilibrium position? ( j) At what times is the mass 5 inches below the equilibrium position? (k) At what times is the mass 5 inches below the equilibrium position heading in the upward direction?

Abdelrahman T.
Answer:

(a) The spring/mass system : free undamped motion has the DE $$ \dfrac{d^2x}{dt^2}+\dfrac{k}{m} x=0 \because \quad m=\dfrac{W}{g}=\dfrac{10 lb}{32 \quad ft/s^2}=\dfrac{10}{32} Slug \qquad$$ and $$\qquad F=W=kx \to\quad 64 lb=k(0.32 ft)\quad\to\quad k=\dfrac{64}{0.32} \quad\to\quad k=200 lb/ft $$ Now, substitute in the DE to get $$\dfrac{d^2x}{dt^2}+\dfrac{200}{2} x=0\qquad \to \qquad \dfrac{d^2x}{dt^2}+100\, x=0 $$ Thus, the auxiliary equation is $$ m^2+100=0 \to\quad m^2=-100\quad\to\quad m=\pm\sqrt{-100}\quad\to\quad m_1=10\, i \qquad and \qquad m_2=-10\, i \to\quad \alpha =0\qquad and \qquad \beta =10 $$ From equation (8) of section 3.2 we obtain the general solution $$ x(t)=t^{\alpha}(c_1\,\cos \beta t+c_2\,\sin \beta t)\quad\to\quad x(t)=c_1\,\cos 10t+c_2\,\sin 10t \to\quad x'(t)=-10\,c_1\,\sin 10\,t+10\,c_2\,\cos 10\,t $$ Since the mass was initially released from a point 8 in above the equilibrium position with downward velocity of $$ 5 \quad ft/s \quad \because \quad 8 in =\dfrac{8}{12} ft=\dfrac{2}{3} ft \qquad \to \quad x(0)=-\dfrac{2}{3} \qquad and \qquad x'(0)=5 $$ Now, applying the initial condition $$ x(0)=-\dfrac{2}{3} $$ we get $$ -\dfrac{2}{3}=c_1\,\cos (0)+c_2\,\sin (0)\quad\to\quad c_1+\,c_2\,(0)=-\dfrac{2}{3} \quad\to\quad c_1=-\dfrac{2}{3} $$ Then, applying the second initial condition $$ x'(0)=5 $$we get $$ 5=-10\,c_1\,\sin (0)+10\,c_2\,\cos (0)\quad\to\quad -10\,c_1\,(0)+10\,c_2=5 \to\quad 10\,c_2=5\quad\to\quad c_2=\dfrac{1}{2}$$ Now, substitute by these results in the general solution to get $$ x(t)=-\dfrac{2}{3} \cos 10t+\dfrac{1}{2} \sin 10t $$ (b) Using equations (6) and (7) we get $$ A=\sqrt{c_1^2+c_2^2}\quad\to\quad A=\sqrt{\left (-\dfrac{2}{3}\right )^2 +\left(\dfrac{1}{2}\right)^2} \quad\to\quad A=\sqrt{\dfrac{4}{9}+\dfrac{1}{4} \quad\to\quad A=\sqrt{\dfrac{25}{36}}\quad\to\quad A=\dfrac{5}{6} ft \quad\to\quad \phi =\tan^{-1} \left (\dfrac{c_1}{c_2} \right)} \quad\to\quad \phi =\tan^{-1} \left (\dfrac{-2/3}{1/2}\right )\quad\to\quad \phi =\tan^{-1} \left (-\dfrac{4}{3}\right )=-0.927 \quad\because\quad \sin \phi > 0 \qquad$$ and $$\qquad \cos \phi < 0 \quad\to\quad \phi $$ lies in the fourth quadrant $$ \qquad \quad\to\quad \phi =-0.927 rad \quad\to\quad x (t)=A\,\sin (\omega t+\phi )\quad\to\quad x (t)=\dfrac{5}{6} \sin (10t-0.927) \quad\because\quad T=\dfrac{2\pi}{\omega}\quad\to\quad T=\dfrac{2\pi}{10}\quad\to\quad T=\dfrac{\pi}{5} seconds $$ (c) After $$ 3\pi $$ seconds the mass will have completed Cycles$$ =\dfrac{t}{T}=\dfrac{3\pi}{\pi/5}\quad\to\quad Cycles=15 $$ (d) For equilibrium position $$ x=0 , $$and the first time the mass passes through equilibrium position at $$ \sin (0) $$ Thus, for the second time we have $$ \sin (2\pi)=0 \quad\to\quad 10t-0.927=2\pi\quad\to\quad 10t=2\pi+0.927\quad\to\quad 10t=7.21 \quad\to\quad t=\dfrac{7.21}{10}\quad\to\quad t=0.721 \quad seconds $$ (e) At the extreme displacement, the velocity = 0 $$ \quad\to\quad x'=0\qquad\qquad \quad\because\quad x'=\dfrac{25}{3} \cos (10t-0.927) \quad\to\quad \dfrac{25}{3} \cos (10t-0.927)=0\quad\to\quad \cos (10t-0.927)=0 \quad\to\quad 10t-0.927=\cos^{-1} 0\quad\to\quad 10t-0.927=n\pi+\dfrac{\pi}{2} \quad\to\quad 10t=n\pi+\dfrac{\pi}{2}+0.927 \quad\to\quad t=\dfrac{1}{10} \left (n\pi+\dfrac{\pi}{2}+0.927\right) \quad\to\quad t=\dfrac{1}{10} \left (\dfrac{(2n+1)\pi}{2}+0.927\right) \quad\to\quad t=\dfrac{(2n+1)\pi}{20}+0.0927 \qquad n=0,1,2,... $$ (f) Now, substitute by $$ t=3 $$ in the solution to get $$ x(3)=\dfrac{5}{6} \sin (10(3)-0.927)\quad\to\quad x(3)=\dfrac{5}{6} \sin (30-0.927) \quad\to\quad x(3)=\dfrac{5}{6} \sin (29.073)\quad\to\quad x(3)=\dfrac{5}{6} \times (-0.716) \quad\to\quad x(3)=-0.597 \quad ft $$ Negative sign means above equilibrium position (g) Now, substitute by $$ t=3 $$ in the equation of $$ x'(t) $$ to get $$ x'(3)=\dfrac{25}{3} \cos (10(3)-0.927)\quad\to\quad x'(3)=\dfrac{25}{3} \cos (30-0.927) \quad\to\quad x'(3)=\dfrac{25}{3} \cos (29.073)\quad\to\quad x'(3)=\dfrac{25}{3} \times (-0.698) \quad\to\quad x'(3)=-5.814 \quad ft/s $$ Negative sign means in the upward direction (h) $$ \quad\because\quad x''=\dfrac{d}{dx} (x')\quad\to\quad x''=\dfrac{d}{dx} \left [\dfrac{25}{3} \cos (10t-0.927)\right] \to\quad x''=-\dfrac{250}{3} \sin (10t-0.927)$$ Now, substitute by $$ t=3 $$ in the equation of $$ x''(t) $$to get $$x''(3)=-\dfrac{250}{3} \sin (10(3)-0.927)\quad\to\quad x''(3)=-\dfrac{250}{3} \sin (30-0.927) \quad\to x''(3)=-\dfrac{250}{3} \sin (29.073)\quad\to\quad x''(3)=-\dfrac{250}{3} \times (-0.716) \quad\to\quad x''(3)=59.702 ft/s^2 $$ Positive sign means in the downward direction. (i) when mass passes through the equilibrium position we have $$ 10t-0.927=0 \quad or \quad 2n\pi $$ Now, substitute by this result in the equation of $$ x'(t) $$ to get $$ x'=\dfrac{25}{3} \cos (2n\pi)\quad\to\quad x'=\dfrac{25}{3} (\pm 1)\quad\to\quad x'=\pm 8.33 \quad ft/s $$ (j) Now, substitute by $$ x=\dfrac{5}{12} $$ in the solution to get $$ \dfrac{5}{12}=\dfrac{5}{6} \sin (10t-0.927)\quad\to\quad \dfrac{1}{2}=\sin (10t-0.927) \quad\to\quad 10t-0.927=\sin^{-1} \left (\dfrac{1}{2}\right)\quad\to\quad 10t-0.927=\dfrac{\pi}{6}+2n\pi \quad\to\quad 10t=0.927+\dfrac{\pi}{6}+2n\pi \quad\to\quad t =\dfrac{1}{10} \left (\dfrac{\pi}{6}+2n\pi+0.927\right ) \quad$$ and $$\quad t=\dfrac{1}{10} \left (\dfrac{5\pi}{6}+2n\pi+0.927\right )\qquad\quad n=0,1,2,... $$ (k) Now, we choose from part (j) the times for which $$ x'< 0 \quad\to\quad t=\dfrac{1}{10} \left (\dfrac{5\pi}{6}+2n\pi+0.927\right )\qquad\quad n=0,1,2,...$$

Algebra
TutorMe
Question:

A customer deposits $2000 in a savings account that pays interest compounded quarterly. How much money will the customer have in the account after 2 yr? After 5 yr?

Abdelrahman T.
Answer:

You can use the following formula to find the balance of an account that earns compound interest. $$A = P\,(1+ \frac{r}{n})^{nt}$$ , where: A = the balance P = the principal (the initial deposit) r = the annual interest rate (expressed as a decimal) n = the number of times interest is compounded per year t = the time in years Interest = $$5.2\% = 0.052$$ Therefore, after 2 years, $$A = 2000(1+\frac{0.052}{4})^{4×2} = $2217.71$$ And, after 5 years, $$A = 2000(1+\frac{0.052}{4})^{4×5} = $2589.52$$

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