help please it's weird question

Question

The rate of rainfall during a storm t hours after it began was 3 mm per hour when t = 5, 6 mm per hour when t = 9 and 5 mm per hour when t = 13. Assuming that a quadratic model applies, find an expression for the rate of rainfall, r mm per hour, in terms of t.

Answer 1

Let r be the rate of rainfall, in mm per hour, and let t be the time, in hours, after the storm began. We can assume a quadratic model of the form r = at^2 + bt + c, where a, b, and c are constants to be determined.

We know that when t = 5, r = 3, so we have the equation: 3 = 25a + 5b + c

When t = 9, r = 6, so we have the equation: 6 = 81a + 9b + c

When t = 13, r = 5, so we have the equation: 5 = 169a + 13b + c

We now have three equations in three variables, which we can solve using standard methods of linear algebra. One way to do this is to eliminate c from the equations by subtracting the first equation from the other two, which gives:

3a + 4b = 3 88a + 8b = 3

Solving this system of linear equations for a and b, we get: a = -1/44 b = 45/22

Substituting these values for a and b in any of the three equations for r, we can solve for c. For example, using the first equation, we have:

3 = 25(-1/44) + 5(45/22) + c c = 9/4

Therefore, the expression for the rate of rainfall as a function of time is: r = (-1/44)t^2 + (45/22)t + 9/4

To find the time at which the rate of rainfall is maximum (i.e., the time of peak rainfall), we can take the derivative of this expression with respect to t and set it equal to zero:

dr/dt = (-2/44)t + 45/22 = 0

Solving for t, we get: t = 495/44

This is the time, in hours, after the storm began when the rate of rainfall is maximum. To find the maximum rate of rainfall, we can substitute this value of t into the expression for r:

r(495/44) = (-1/44)(495/44)^2 + (45/22)(495/44) + 9/4 r(495/44) = 495/88 + 10125/968 - 198/44 r(495/44) = 25581/968 r(495/44) ≈ 26.43

Therefore, the maximum rate of rainfall during the storm is approximately 26.43 mm per hour.

Answer 2

m =-(t^2)/8 +5t/2 -51/8

is the standard quadratic equation graphed as a curve through the given points (t,m) = (5,3), (9,6) and (13,5)

where t = hours and m = rain milllimeters per hour

in standard general form:

y=ax^2 +bx +c where

a= -1/8

b = 5/2

c =-51/8

the maximum rainfall is the vertex of the parabola (10, 6.125)= (10, 6 1/8) = maximum point

in 10 hours, rain fall = 6.125 mm per hour

just from looking at a rough sketch of the graph, you can tell the maximum is only slightly more than 6, just trying to fit a downward opening parabola through the 3 points.

plug in these 3 points (5,3), (9,6), (13,5) to check the solution, the quadratic equation

m(t) =-(t^2)/8 +(5/2)t - 51/8

3 = -25/8 +25/2 -51/8 = (-25+100-51)/8 = 24/8= 3

6 = -81/8 + 45/2 -51/8 = (-81 +180-51)/8 = 48/8 = 6

5 = -169/8 +65/2 -51/8 = (-169+260-51)/8 = 40/8=5

t mm

5 3

9 6

13 5

graph t on the x axis, mm on the vertical y axis

m = a(t-h)^2+k where (h,k) = vertex=max for a downward opening parabola

3 = a(5-h)^2+k

6= a(9-h)^2 +k

5 =a(13-h)^2 + k

3 equations, 3 unknowns, solve for a,h and k

by substitution, elimination or matrix algebra

h=10 hours

k = 6.125 mm per hour

a = -1/8 = -0.125

or find a horizontally leftward opening parabola, but it's not a function as it fails the vertical line test

only the vertical downward opening parabola is a function

this is the standard form for a horizontal parabola

t = a(m-k)^2 +h

5= a(3-k)^2 + h

9= a(6-k)^2 + h

13 = a(5-k)^2 + h

three equations, three unknowns, solve for a, k and h where (k,h) is the vertex

a= -8/3

b= 76/3

c =-47

for t = am^2 +bm +c

or

a = -8/3

h =4.75

k = 14 5/6

for t = a(m-k)^2 +h in vertex form

vertex is the most rightward point on the parabola ( 14 5/6, 4 3/8)

with maximum number of hours = about 14.67 hrs

and rainfall about 4.13 mm per hr

but for any given number of hours there are two possible rain fall rates (other than at the vertex)

maximum rainfall rate is no longer at the vertex but when t = 0 hours and even then it could also be the minimum rainfall, as there are two mm values for any one t value

at t = 0, 0 = -(8/3)m^2 + 76x/3 -47 which gives 2 solutions

using the quadratic formula, one a max, the other a minimum rainfall rate

the vertex gives the maximum number of hours of the storm= 14 5/6 hours

there are an infinite number of possible parabolas at different angles, other than vertical or horizontal

but the most standard is the vertical, then 2nd place the horizontal opening parabola

You want the 1st solution, for the downward opening vertical parabola

it may help to plot the points and try to fit a parabola through them

even just a rough sketch, visually.

the above calculations can be very tedious, with mistakes easily made, in just arithmetic. But the above numbers pan out. Check them, they're correct.