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3 years ago

What is the rectangular form of the polar equation?

Mathematics

1 answer:

lianna [129]3 years ago

3 0

Answer:

Option (1)

Step-by-step explanation:

From the picture attached,

tanθ = $\frac{y}{x}$

Given : Polar equation as 'θ' = $-\frac{5\pi }{6}$

Therefore, $\text{tan}(-\frac{5\pi }{6} )$ = $\frac{y}{x}$

$-\text{tan}(\frac{5\pi }{6} )$ = $\frac{y}{x}$ [Since tan(-θ) = -tanθ]

$\text{tan}(\pi -\frac{5\pi }{6} )$ = $\frac{y}{x}$ [Since -tanθ = tan(π - θ)]

$\text{tan}\frac{\pi }{6}$ = $\frac{y}{x}$

$\frac{y}{x}=\frac{\sqrt{3}}{3}$

y = $\frac{\sqrt{3} }{3}x$

Therefore, y = $\frac{\sqrt{3} }{3}x$ will be the rectangular form of the polar equation.

Option (1) will be the correct option.

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4. Which of the following sets of ordered pairs is a function?

alisha [4.7K]

Answer:

Step-by-step explanation:

Thing that x = a kid and y = it's biological mother

To be a function you need for each x to have only one y !

a. {(x= -3, y = -2), (x= -1, y= 3), (x=0, y= -2), (x= 3, y= 4)} Is a Function

b. {(x= 0, y= 1), (x= 3, y= 2), (x= 5, y= -3), (x= 0, y= 2)} Not a Function because kid x=0 has two biological mothers y=1 and y=2.

c. {(x= -7, y= -7) (x= -2, y= 5),(x= -1, y= 6).(x= -2, y= -5)}Not a Function because kid x= -2 has two biological mothers y= 5 and y= -5

d. {(x= -4, y= -7), (x= -9, y= 5), (x= 4, y= -2), (x= -9, y= 0)} Not a Function because kid x= -9 has two biological mothers y= 5 , and y= 0

6 0

3 years ago

Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−

aleksandrvk [35]

The tangent to $C$ through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces $S$ and $T$ at that point.

Let $f(x,y,z)=x^2+2y^3+3z^4$ . Then $S$ is the level curve $f(x,y,z)=6$ . Recall that the gradient vector is perpendicular to level curves; we have

$\nabla f(x,y,z)=(2x,6y,12z^2)$

so that the gradient of $f$ at (1, 1, 1) is

$\nabla f(1,1,1)=(2,6,12)$

For the surface $T$ , we have

$\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0$

so that $\vec r(1,0)=(1,1,1)$ . We can obtain a vector normal to $T$ by taking the cross product of the partial derivatives of $\vec r(u,v)$ , and evaluating that product for $u=1,v=0$ :

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)$

$\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)$

Now take the cross product of the two normal vectors to $S$ and $T$ :

$(2,6,12)\times(1,-1,2)=(24,8,-8)$

The direction of vector (24, 8, -8) is the direction of the tangent line to $C$ at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by $t\in\Bbb R$ . Then adding (1, 1, 1) shifts this line to the point of tangency on $C$ . So the tangent line has equation