Can someone please help me

Mathematics

2 answers:

Misha Larkins [42]3 years ago

6 0

Try 9cm but I’m not quite sure please reply if it’s wrong or right

Alja [10]3 years ago

5 0

I agree with Humairak05. try 9 I tried to the math

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mario62 [17]

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ivann1987 [24]

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2 years ago

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kap26 [50]

Step-by-step explanation:

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2 years ago

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3 years ago

If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$

6 0

3 years ago