In meteorology, rainfall measurement and forecast are often encountered.
Typhoon, sandstorm, movement law of cold current center and prediction of water level rise
Such problems are often transformed into mathematical problems to solve.
Now, for example.
First, measure the rainfall.
The rainfall of 1 refers to the rainfall per unit area on the horizontal ground.
Depth of rain. Now the upper opening has a diameter of 32cm, and the bottom is straight.
A frustum barrel with a diameter of 24 cm and a depth of 35 cm is used to measure the drop.
Rainfall If it rains, the rain in this bucket
If it is a quarter of the barrel depth, what is the rainfall?
Millimeter (accurate to 1 mm)
Analysis: To require rainfall, just calculate the unit area.
The depth of rain, and the depth of rain per unit area, can
Solve by equal product.
Solution: According to the meaning of the question, the water depth of the frustum-shaped bucket is O 1O2=354cm, and because a1b1a1= aba2b.
, so a1b1= aba2b1a2b = (16-12) * 35435 =1,so the water surface.
The radius o1a1=12+1=13 (cm), so the rainwater is in the bucket.
The volume is v water = 1
3π( 122+ 12× 13+ 132)×354 = 164 15 12π(cm)。
Because the area of the upper mouth of the bucket is
On s =π 162= 256π(cm2), let each
1cm2 rainfall is xcm, then
X = V s on the water =16415π121256π ≈ 513 (cm).
Therefore, the rainfall is about 53 mm.
Note: In addition to defining the concept of rainfall,
It is also necessary to deeply understand the meaning of the question and get the calculation method of rainfall.
Method. Why divide the volume of rainwater by the area of the barrel mouth, and
Not divided by water surface area or other areas.
It is difficult to analyze and reason. In fact, in the process of rainfall, rain
Water falls into the barrel mouth, so it can hold more rain.
At least it has something to do with the size of the barrel mouth, not with the shape of the barrel itself.
It is not difficult to understand the above method of calculating rainfall.
Second, the typhoon forecast
According to the forecast of the Meteorological Observatory, it is 300 kilometers east of S Island.
A There is a typhoon center, 40 per hour.
The speed of kilometers moves to the northwest, 250 meters away from the typhoon center.
Areas within kilometers will be affected by it. Q: It has passed since then.
How long will the typhoon affect S Island and how long will it last?
Analysis: the typhoon center is moving, and its moving law is as follows
What? We can set up a coordinate system to study this.
problem Take S Island as the origin, as shown in Figure 2.
Establish a plane rectangular coordinate system as shown in.
X Sy, then the coordinates of point a are
(300,0), the equation of circle S is x2+
Y2= 2502。 It is easy to know where the typhoon center is.
When the circle s is above or inside, the typhoon will have a shadow.
Around s island, know that the typhoon center is 40km/h.
Move to the northwest, so that the ray can be set at the location of the typhoon center.
The parameter equation of l is
x= 300 + 40tcos 135，
y= 40tsin 135
(t≥0)，
Among them, the physical meaning of parameter t is time (hours).
So the question becomes "When the time t is in what range,
The typhoon center is inside or on the boundary of circle S.
Solution: We assume the trajectory of the typhoon central ray L.
The parameter equation is
x= 300 + 40tcos 135，
y= 40tsin 135
(t≥0), that is, in a typhoon.
The heart is (300-202t, 202t).
Therefore, the necessary and sufficient conditions for the typhoon center to be on or in the circle.
be
(300 - 202t)2+(202t)2≤2502，
The solution method is1199 ≤ t ≤ 8161.
04 middle school mathematics teaching reference number1~ 2,2001
So in about two hours, S Island will be affected by the typhoon.
It lasted about 6 16 hours.
Description: This topic is very important for studying typhoon, sandstorm and cold current center.
The influence of movement law on guiding and preventing natural disasters is realistic.
Meaning.
Third, predict the water level rise.
There is a reservoir somewhere, which is the most important when it is built.
The design of large water quantity is 128000m3. In case of flash floods, plan ahead.
Measure the amount of water Sn (unit: m3) injected into the reservoir and the number of days n(n
∈N, n≤ 10), then Sn= 5000.
N(n+ 24)。 The original water volume of this reservoir is 80000 cubic meters.
The daily discharge of the sluice is 4000 cubic meters. In case of flash floods,
On the first day, I opened the sluice and asked: Has there been any dam break in this 10 day?
Danger (when the reservoir water exceeds the maximum amount, the dam will burst)
Analysis: This is a modeling element about unreasonable inequality.
Material, the following mathematical model can be established: 5000n(n+ 24)
-4000n & gt； 128000-80000, the solution is n >；; 8. Reservoir
The dam will be in danger on the ninth day.
Due to the arrival of flood peak, under a parabolic arch bridge.
A rescue boat was ordered to swim 8 kilometers away and asked to start at once.
Go to the upper reaches of the bridge to perform the task, and inform the current speed at this time.
100m/min, the water span of the arch bridge is 30m, and the water surface is
Upper arch height10m, rising height and time t of water surface under the bridge.
Is proportional to the square of (minutes), and the proportional coefficient is
1
1000
. known
The width and height of the rescue boat's surface are 3 meters, so I ask this question.
At least how fast can the ship go before it can pass smoothly.
The speed is considered to be uniform)
Analysis: To make the ship pass smoothly, as long as the bridge arches to the water.
The width at 3 meters above the water is greater than or equal to the width of the ship.
Solution: Establish a straight line as shown in Figure 3.
Angle coordinate system of parabolic arch bridge
The equation is y =-ax2 (a >; 0).
Will point A(302
,-10) into parabolic equation, we can get a=43.
So the equation of parabola is y= -43x2.
When the ship reaches the bridge mouth within t minutes, the width of the ship is positive.
Good is equal to the span of the bridge arch 3 meters above the water surface, and the ship is just right at this time.
So that we can cross the bridge. Therefore, the water level under the bridge rises 1 1000t2m, leaving the bridge.
The coordinate of point B on the arch curve, which is 3m away from the water surface, is (32,-10+3+ 1000t2). If we substitute the parabolic equation, we can get-7+11000t2 =-43x (32) 2, that is.
(minutes), so, in order to let the ship pass smoothly, it must be used.
The time is less than or equal to 20 10 minutes, so as to set the ship speed.
Degree is V (m/min), then
8000
V- 100≤20 10, that is, v≥8000.
2010+100 = 22615 (m/min), so the speed of the ship.
At least 226 15 m/min can pass smoothly.
Explanation: The key to solve this problem is to solve it by parabolic equation first.
Find its time t, and then solve the inequality about speed v.