Answer: im pretty sure its 5 to the second power so the third one
Step-by-step explanation:
(a)
![\displaystyle \frac{1}{2} \cdot \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%20%5Cint_%7B%5Cfrac%7B%5Cpi%7D%7B3%7D%7D%5E%7B%5Cfrac%7B5%5Cpi%7D%7B3%7D%7D%20%5Cleft%284%5E2%20-%20%283%20%2B%202%5Ccos%5Ctheta%29%5E2%20%5Cright%29%20%5C%2C%20d%5Ctheta)
or, via symmetry
![\displaystyle\frac{1}{2} \cdot 2 \int_{\frac{\pi}{3}}^{\pi} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%202%20%5Cint_%7B%5Cfrac%7B%5Cpi%7D%7B3%7D%7D%5E%7B%5Cpi%7D%20%5Cleft%284%5E2%20-%20%283%20%2B%202%5Ccos%5Ctheta%29%5E2%20%5Cright%29%20%5C%2C%20d%5Ctheta)
____________
(b)
By the chain rule:
![\displaystyle \frac{dy}{dx} = \frac{ dy/ d\theta}{ dx/ d\theta}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7B%20dy%2F%20d%5Ctheta%7D%7B%20dx%2F%20d%5Ctheta%7D)
For polar coordinates, x = rcosθ and y = rsinθ. Since
<span>r = 3 + 2cosθ, it follows that
![x = (3 + 2\cos\theta) \cos \theta \\ y = (3 + 2\cos\theta) \sin \theta](https://tex.z-dn.net/?f=x%20%3D%20%283%20%2B%202%5Ccos%5Ctheta%29%20%5Ccos%20%5Ctheta%20%5C%5C%20%0Ay%20%3D%20%283%20%2B%202%5Ccos%5Ctheta%29%20%5Csin%20%5Ctheta)
Differentiating with respect to theta
![\begin{aligned} \displaystyle \frac{dy}{dx} &= \frac{ dy/ d\theta}{ dx/ d\theta} \\ &= \frac{(3 + 2\cos\theta)(\cos\theta) + (-2\sin\theta)(\sin\theta)}{(3 + 2\cos\theta)(-\sin\theta) + (-2\sin\theta)(\cos\theta)} \\ \\ \left.\frac{dy}{dx}\right_{\theta = \frac{\pi}{2}} &= 2/3 \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%0A%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%26%3D%20%5Cfrac%7B%20dy%2F%20d%5Ctheta%7D%7B%20dx%2F%20d%5Ctheta%7D%20%5C%5C%0A%26%3D%20%5Cfrac%7B%283%20%2B%202%5Ccos%5Ctheta%29%28%5Ccos%5Ctheta%29%20%2B%20%28-2%5Csin%5Ctheta%29%28%5Csin%5Ctheta%29%7D%7B%283%20%2B%202%5Ccos%5Ctheta%29%28-%5Csin%5Ctheta%29%20%2B%20%28-2%5Csin%5Ctheta%29%28%5Ccos%5Ctheta%29%7D%20%5C%5C%20%5C%5C%0A%5Cleft.%5Cfrac%7Bdy%7D%7Bdx%7D%5Cright_%7B%5Ctheta%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%0A%26%3D%202%2F3%0A%5Cend%7Baligned%7D)
2/3 is the slope
____________
(c)
"</span><span>distance between the particle and the origin increases at a constant rate of 3 units per second" implies dr/dt = 3
A</span>ngle θ and r are related via <span>r = 3 + 2cosθ, so implicitly differentiating with respect to time
</span><span />
Answer: 4:3.
Step-by-step explanation:
Given: Point P is
of the distance from M to N.
To find: The ratio in which the point P partition the directed line segment from M to N.
If Point P is between points M and N, then the ratio can be written as
![\dfrac{MP}{MN}=\dfrac{MP}{MP+PN}](https://tex.z-dn.net/?f=%5Cdfrac%7BMP%7D%7BMN%7D%3D%5Cdfrac%7BMP%7D%7BMP%2BPN%7D)
As per given,
![\dfrac{MP}{MP+PN}=\dfrac{4}{7}\\\\\Rightarrow\ \dfrac{MP+PN}{MP}=\dfrac{7}{4}\\\\\Rightarrow\ \dfrac{MP}{MP}+\dfrac{PN}{MP}=\dfrac{7}{4}\\\\\Rightarrow\ -1+\dfrac{PN}{MP}=\dfrac{7}{4}\\\\\Rightarrow\ \dfrac{PN}{MP}=\dfrac{7}{4}-1=\dfrac{7-4}{4}=\dfrac{3}{4}\\\\\Rightarrow\ \dfrac{PN}{MP}=\dfrac{3}{4}\ \ \or\ \dfrac{MP}{PN}=\dfrac{4}{3}](https://tex.z-dn.net/?f=%5Cdfrac%7BMP%7D%7BMP%2BPN%7D%3D%5Cdfrac%7B4%7D%7B7%7D%5C%5C%5C%5C%5CRightarrow%5C%20%5Cdfrac%7BMP%2BPN%7D%7BMP%7D%3D%5Cdfrac%7B7%7D%7B4%7D%5C%5C%5C%5C%5CRightarrow%5C%20%5Cdfrac%7BMP%7D%7BMP%7D%2B%5Cdfrac%7BPN%7D%7BMP%7D%3D%5Cdfrac%7B7%7D%7B4%7D%5C%5C%5C%5C%5CRightarrow%5C%20-1%2B%5Cdfrac%7BPN%7D%7BMP%7D%3D%5Cdfrac%7B7%7D%7B4%7D%5C%5C%5C%5C%5CRightarrow%5C%20%5Cdfrac%7BPN%7D%7BMP%7D%3D%5Cdfrac%7B7%7D%7B4%7D-1%3D%5Cdfrac%7B7-4%7D%7B4%7D%3D%5Cdfrac%7B3%7D%7B4%7D%5C%5C%5C%5C%5CRightarrow%5C%20%5Cdfrac%7BPN%7D%7BMP%7D%3D%5Cdfrac%7B3%7D%7B4%7D%5C%20%5C%20%5Cor%5C%20%5Cdfrac%7BMP%7D%7BPN%7D%3D%5Cdfrac%7B4%7D%7B3%7D)
Hence, P partition the directed line segment from M to N in 4:3.
Answer:
4.56
Step-by-step explanation: