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jeka94
3 years ago
8

Part b only please if you have time

Physics
1 answer:
andrew-mc [135]3 years ago
7 0
To convert parametric to Cartesian systems, you need to find a way to get rid of the t's.

In this case, the t's are inside trigonometric functions, so we're going to use a very famous trig identity you should memorize:

{sin(t)}^{2} + {cos(t)}^{2} = 1

If we plug sin(t) and cos(t) into that equation only x and y variables will be left!

BUT there's one thing. The given cos(t + pi/6) has nasty extra stuff in it. However, part a gives you a tip on how to relate x and y to a nice clean cos(t)

So if we do a little rearranging:

\sin(t) = \frac{y}{2} \\ \cos(t) = \frac{x + y}{2 \sqrt{3} }

Now we can plug these into the famous trig identity!

{( \frac{y}{2}) }^{2} + {( \frac{x + y}{2 \sqrt{3} } )}^{2} = 1

Do a little bit of adjustments to get that final form asked for, and you'll be able to find those integers of a and b. ;)
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Observe that the given vector field is a gradient field:

Let f(x,y,z)=\nabla g(x,y,z), so that

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Integrating the first equation with respect to x, we get

g(x,y,z) = \dfrac12 x^2 y^2 z^2 + h(y,z)

Differentiating this with respect to y gives

\dfrac{\partial g}{\partial y} = x^2 y z^2 + \dfrac{\partial h}{\partial y} = x^2 y z^2 \\\\ \implies \dfrac{\partial h}{\partial y} = 0 \implies h(y,z) = i(z)

Now differentiating g with respect to z gives

\dfrac{\partial g}{\partial z} = x^2 y^2 z + \dfrac{di}{dz} = x^2 y^2 z \\\\ \implies \dfrac{di}{dz} = 0 \implies i(z) = C

Putting everything together, we find a scalar potential function whose gradient is f,

f(x,y,z) = \nabla \left(\dfrac12 x^2 y^2 z^2 + C\right)

It follows that the curl of f is 0 (i.e. the zero vector).

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