The average sunflower has 34 petals. What is the best estimate of the total number of petals on 57 sunflowers?

1 answer:

3 0

Your initial instinct would be to solve this problem with this solution. 34 x 57 = 1,938 petals.

However, the word estimate is present; therefore, we have to consider that we need to get the approximate number and not the exact number of petals.

There is a general rule regarding estimation or approximation. You have to round it to the nearest place you have based on. In this problem we round it off to the nearest tens place.

If the number is below 5, you round it down: 34 is rounded down to 30 petals.

If the number is above 5, you round it up: 57 is rounded up to 60 sunflowers.

Thus, your equation would be : 30 x 60 = 1,800 petals (best estimate)

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Find the surface area of x^2+y^2+z^2=9 that lies above the cone z= sqrt(x^@+y^2)

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The cone equation gives

$z=\sqrt{x^2+y^2}\implies z^2=x^2+y^2$

which means that the intersection of the cone and sphere occurs at

$x^2+y^2+(x^2+y^2)=9\implies x^2+y^2=\dfrac92$

i.e. along the vertical cylinder of radius $\dfrac3{\sqrt2}$ when $z=\dfrac3{\sqrt2}$ .

We can parameterize the spherical cap in spherical coordinates by

$\mathbf r(\theta,\varphi)=\langle3\cos\theta\sin\varphi,3\sin\theta\sin\varphi,3\cos\varphi\right\rangle$

where $0\le\theta\le2\pi$ and $0\le\varphi\le\dfrac\pi4$ , which follows from the fact that the radius of the sphere is 3 and the height at which the sphere and cone intersect is $\dfrac3{\sqrt2}$ . So the angle between the vertical line through the origin and any line through the origin normal to the sphere along the cone's surface is

$\varphi=\cos^{-1}\left(\dfrac{\frac3{\sqrt2}}3\right)=\cos^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4$

Now the surface area of the cap is given by the surface integral,

$\displaystyle\iint_{\text{cap}}\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dv\,\mathrm du$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}9\sin v\,\mathrm dv\,\mathrm du$
$=-18\pi\cos v\bigg|_{v=0}^{v=\pi/4}$
$=18\pi\left(1-\dfrac1{\sqrt2}\right)$
$=9(2-\sqrt2)\pi$