(-6u+3c+4)+(5u-5c) help me please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

alexira [117]4 years ago

6 0

(-6u + 3c + 4) + (5u - 5c)
-1u - 2c + 4

tigry1 [53]4 years ago

5 0

-6u+3c+4+5u-5c
-1u-2c+4

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Find the center of mass of the wire that lies along the curve r and has density =4(1 sin4tcos4t)

dolphi86 [110]

The mass of the wire is found to be 40π√2 units.

To calculate the mass of the wire which runs along the curve r ( t ) with the density function δ=5.

The general formula is,

Mass = $\int_a^b \delta\left|r^{\prime}(t)\right| d t$

To find, we must differentiate this same given curve r ( t ) with respect to t to estimate |r'(t)|.

The given integration limits in this case are a = 0, b = 2π.

Now, as per the question;

The equation of the curve is given as;

r(t) = (4cost)i + (4sint)j + 4tk

Now, differentiate this same given curve r ( t ) with respect to t.

$\begin{aligned}\left|r^{\prime}(t)\right| &=\sqrt{(-4 \sin t)^2+(4 \cos t)^2+4^2} \\&=\sqrt{16 \sin ^2 t+16 \cos ^2 t+16} \\&=\sqrt{16\left(\sin t^2+\cos ^2 t\right)+16}\end{aligned}$

Further simplifying;

$\begin{aligned}&=\sqrt{16(1)+16} \\&=\sqrt{16+16} \\&=\sqrt{32} \\\left|r^{\prime}(t)\right| &=4 \sqrt{2}\end{aligned}$

Now, use integration to find the mass of the wire;

$\begin{aligned}&=\int_a^b \delta\left|r^{\prime}(t)\right| d t \\&=\int_0^{2 \pi} 54 \sqrt{2} d t \\&=20 \sqrt{2} \int_0^{2 \pi} d t \\&=20 \sqrt{2}[t]_0^{2 \pi} \\&=20 \sqrt{2}[2 \pi-0] \\&=40 \pi \sqrt{2}\end{aligned}$

Therefore, the mass of the wire is estimated as 40π√2 units.

To know more about density function, here

brainly.com/question/27846146

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The complete question is-

Find the mass of the wire that lies along the curve r and has density δ.

r(t) = (4cost)i + (4sint)j + 4tk, 0≤t≤2π; δ=5

5 0

2 years ago

PLEAS HELP ME

GarryVolchara [31]

$\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{Table~A}{\begin{array}{ccll} x&y\\ \cline{1-2} 60&48\\70&56\\80&64 \end{array}}~\hspace{7em} \stackrel{Table~B}{\begin{array}{ccll} x&y\\ \cline{1-2} 20&24\\30&36\\40&48 \end{array}} \\\\[-0.35em] ~\dotfill$

$\bf \textit{for table A, we know that } \begin{cases} x=60\\ y=48 \end{cases}\implies 48=k60\implies \cfrac{48}{60}=k \\\\\\ \cfrac{4}{5}=k\qquad therefore\qquad \boxed{y=\cfrac{4}{5}x} \\\\\\ \textit{for table B, we know that } \begin{cases} x=30\\ y=36 \end{cases}\implies 36=k30\implies \cfrac{36}{30}=k \\\\\\ \cfrac{6}{5}=k\qquad therefore\qquad \boxed{y=\cfrac{6}{5}x}$