Question

1. Fie o miscare definita prin ecuatiile parametrice:

x(t) = 3 la t + 1/4

y(t) = 4 la t − 1

Sa se determine ecuatia traiectoriei mobilului, viteza si acceleratia sa. Sa se calculeze, de asemenea, modulul vitezei si acceleratiei. Observatie: ecuatiile paramatrice arata cum variaza in functie de timpul t coordonatele x, y si z ce descriu miscarea mobilului. Ecuatia traiectoriei se obtine prin eliminarea timpului din ecuatiile parametrice.


Va multumesc :)!

Answer 1

$x(t) = 3^t + \frac{1}{4}\\ x - \frac{1}{4} = 3^t\\ \ln{(x - \frac{1}{4})} = \ln{3^t}\\\ln{(x-\frac{1}{4})} = t\ln{3}\Rightarrow t = \frac{\ln{ (x-\frac{1}{4} )}}{\ln{3}}\iff \boxed{t = log_3 (x - \frac{1}{4})}$

$y(t) = 4^t - 1\\y+1 = 4^t\\\ln{(y+1)} = \ln{4^t}\\\ln{(y+1)} = t\ln{4}\Rightarrow t = \frac{\ln{(y+1)}}{\ln{4}}\iff \boxed{t = log_4 (y+1)}$

$t = t \iff log_3(x-\frac{1}{4}) = log_4(y+1)\Big | \div log_4(3)\\\frac{log_3(x-\frac{1}{4})}{log_4(3)} = log_3 (y + 1)\\\frac{1}{log_4(3)}\cdot log_3(x-\frac{1}{4}) = log_3 (y+1)\\log_3\Big((x-\frac{1}{4})^{\frac{1}{log_4(3)}}\Big) = log_3(y+1)\Rightarrow y+1 = (x-\frac{1}{4})^{\frac{1}{log_4(3)}}\iff \boxed{y = \sqrt[log_4(3)]{x-\frac{1}{4}}-1}$

Viteza:

$v = \sqrt{v_x^2 + v_y^2}\\\textrm{Viteza pe x este derivata functiei x(t), si viteza pe y este derivata functiei y(t).}\\v_x = \frac{dx}{dt}(t) = [3^t + \frac{1}{4}]' = [3^t]' + [\frac{1}{4}]' = 3^t\ln{3} + 0 = 3^t\ln{3}\\v_y = \frac{dy}{dt}(t) = [4^t-1]' = [4^t]' - [1]' = 4^t\ln{4} - 0 = 4^t\ln{4}$

$v = \sqrt{(3^t\ln{3})^2 + (4^t\ln{4})^2} = \sqrt{3^{2t}(\ln{3})^2 + 4^{2t}(\ln{4})^2}\\\ln{3} = \frac{log_3(3)}{log_3 (e)}= \frac{1}{log_3 (e)}\\\ln{4} = \frac{log_3(4)}{log_3 (e)} \\\\v = \sqrt{3^{2t} \cdot (\frac{1}{log_3 (e)})^2 + 4^2t(\frac{log_3(4)}{log_3(e)})^2} = \sqrt{\frac{1}{(log_3(e))^2} (3^{2t} + 4^{2t}\cdot (log_3(4))^2)} = \frac{\sqrt{3^{2t} + 4^{2t}\cdot (log_3(4))^2}}{log_3(e)} = \frac{\sqrt{9^t + 16^t\cdot (log_3(4))^2}}{log_3(e)}$

Acceleratia:

$a = \frac{dv}{dt} (t) = [\frac{\sqrt{9^t + 16^t\cdot (log_3(4))^2}}{log_3(e)}]' = \frac{1}{log_3(e)}[\sqrt{9^t + 16^t\cdot(log_3(4))^2}]' = \frac{1}{log_3(e)}\cdot \frac{1}{2}\cdot (9^t + 16^t\cdot (log_3(4))^2) \cdot [9^t + 16^t(log_3(4))^2]' = \frac{1}{log_3(e)}\cdot \frac{1}{2}\cdot (9^t + 16^t\cdot (log_3(4))^2) \cdot (9^t\ln{9} + 16^t\ln{16}\cdot(log_3(4))^2) = \frac{1}{log_3(e)}\cdot \frac{1}{2}\cdot (9^t + 16^t\cdot (log_3(4))^2) \cdot (9^t\ln{3}\cdot 2 + 16^t\ln{4}\cdot 2\cdot(log_3(4))^2) = \frac{1}{log_3(e)}\cdot \frac{1}{2}\cdot (9^t + 16^t\cdot (log_3(4))^2) \cdot 2(9^t\ln{3} + 16^t\ln{4}\cdot(log_3(4))^2) = \frac{1}{log_3(e)}\cdot (9^t + 16^t\cdot (log_3(4))^2) \cdot (9^t\ln{3} + 16^t\ln{4}\cdot(log_3(4))^2) = \frac{1}{log_3(e)}({(9^t)}^2\ln{3} + (9\cdot 16)^t\ln{4}\cdot (log_3(4))^2 + (9\cdot 16)^t\ln{3}\cdot (log_3(4))^2 + {(16^t)}^2\ln{4}\cdot (log_3(4))^4)= \frac{1}{log_3(e)}(9^{2t}\ln{3} + (9\cdot 16)^t \cdot (log_3(4))^2 \cdot (\ln{4} + \ln{3}) + 16^{2t}\ln{4}\cdot (log_3(4))^4) = \frac{1}{log_3(e)}(81^t\frac{1}{log_3 (e)} + 144^t(log_3(4))^2\cdot (\frac{log_3(4) + 1}{log_3(e)}) + 256^t\frac{(log_3(4))^5}{log_3(e)}) = \frac{1}{(log_3(e))^2}(81^t + 144^t(log_3(4))^2\cdot (log_3(4) + 1) + 256^t( log_3(4) )^5)$

Daca inlocuim logaritmii cu valorile lor, obtinem:

$v = \frac{\sqrt{9^t + 16^t\cdot (log_3(4))^2}}{log_3(e)} = \frac{\sqrt{9^t + 16^t\cdot 1.59228941577}}{0.91023922662} = 1.09861228867\sqrt{9^t + 16^t\cdot 1.59228941577} = \sqrt{ 1.20694896081\cdot 9^t + 16^t \cdot 1.92181205568}$

$a = \frac{1}{(log_3(e))^2}(81^t + 144^t(log_3(4))^2\cdot (log_3(4) + 1) + 256^t( log_3(4) )^5) = 1.20694896081\cdot (81^t + 144^t \cdot 1.59228941577 \cdot 2.26185950714 + 256^t \cdot 3.19930040289) = 1.20694896081\cdot 81^t + 1.20694896081\cdot 144^t \cdot 3.60153495318 + 1.20694896081 \cdot 256^t\cdot 3.19930040289 = 1.20694896081\cdot 81^t +4.34686886906 \cdot 144^t + 3.86139229659\cdot 256^t$