Question

6道微积分数学题 国外大学的

Answer 1

1、Let the origin be the point where the balls are thrown upward and set a +y direction upward.The first ball has the position equation:y = y0 + v0*t + (1/2)at^2y = 48t - 16t^2The second ball has a position equation:y = y0 + v0*(t - 1) + (1/2)a(t - 1)^2 <-- Since it is launched 1 s later.y = 24(t - 1) - 16(t - 1)^2y = -16t^2 + 56t - 40When do these two y values equal each other? When48t - 16t^2 = -16t^2 + 56t - 40-8t = -40t =5 s
2、Decelrato is just negative acceleration, so we have the car's acceleration: a = -16 The anti-derivative of acceleration is velocity, and the antiderivative of velocity is displacement. So, the function for the velocity of the car is: v = -16t + C In this case, C will be the initial velocity of the car (since when t = 0, v = C ) And we know that at some point the velcity is zero, so we can say: 0 = -16t + C, where time is now the total time it took for the car to stop. Let's start by getting the displacement: s = (-16t^2)/2 + Ct + D, where D is a new constant. = -8t^2 + Ct + D Now, we know that the car's displacement will be zero at time zero, since it can't have moved in zero time. So 0 = -8(0^2) + c(0) + D D = 0 s = -8t^2 + Ct But we also know that the car moves a total of 200 ft before stopping, so: 200 = -8t^2 + Ct Combined with our equation for velocity: 0 = -16t + C We have two equations with two unknowns. If we rearrange the second equation, we get: 16t = C And if we substitute this into the first equation: 200 = -8t^2 + (16t)t 200 = -8t^2 + 16t^2 200 = 8t^2 25 = t^2 5 = t So it took the car 5 seconds to stop. Now we can find C, since C = 16t = 16(5) = 80 So the initial velocity of the car was 80 ft/s
Thenour velocity at t=0 is given by x'= -16(0)+80, which is 80ft/s. Then for the heck of it, our final equations are: x"(t) = -16 x'(t) = -16t + 80 x(t) = -8t^2 +80t

Answer 2

膜拜一下，没考过GRE的估计做不出来。
楼主英文好不如去国外的网站问问。
据我观察考过GRE的都不会关注百度知道的。

Answer 3

dv/dt=a=F/m=-g
v=v0-gt
s=int v dt=v0t-.5gt^2
Assume the height of the cliff is h
then

(1) -h=v01t-.5gt^2
(2) -h=v02(t-1)-.5g(t-1)^2
(1)-(2) 0=48t-24(t-1)-16[t^2-(t-1)^2]
0=24t+24-16(2t-1)
8t=40
t=5
From(1)
h=16t^2-v01t=16*5^2-48*5=400-240=160 ft
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dv/dt=16
v(end)=0
v=v0-16t
t=v0/16
v0=16t
S=int vdt=v0t-8t^2=200
8t^2=200
t=5
v0=16*5=80ft/s
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The increase of the population from time t=0 to t=16
The population at t=16
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Since slope=derivative,so the integral means the increase of the height form x=3 to x=5
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f'(x)=df/dx=pounds/feet
int f(x)dx=pounds*feet
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The points are
pi/n,2pi/n,...,npi/n
The interval is (0,pi)
so h=(pi-0)/n=pi/n
The sum can be rewritten as
int<0,pi> sin(x)dx
=-cosx|<0,pi>
=2