1. A particle moves along the x-axis, it's position at timer given by x(t)=t/(1+t^2), t greater than or equal to 0,where t is measured in seconds and x in meters.
a) find the velocity at time t.
I am a little confused.. Do I find the derivative by using the quotient rule? What would the answer be? I am not sure with mine..
My answer: s'(t)=v(t)=(-t^2 + 1)/(t^4 +2t^2 +1)??
b) when is the particle moving to the right? When is it moving to the left?
Particle moving right when v(t) is greater than zero?? Left when v(t) is less than zero??
c) find the total distance traveled during the first 4s.
??
2.If a ball is thrown vertically upward with a velocity of 80 ft/sec, then it's height after t sec is s=80t-16t^2.
a) what is the maximum height reached by the ball?
Derivative of s =o ????
Is it 2.5 sec?
b) what is the velocity of the ball when it is in 96 ft above the ground on its way up? On its way down?
s(t)=0????
3. If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the volume V of water remaining in the tank after t minutes as
V=5000(1-t/40)^2
t is greater than or equal to 0 and less than or equal to 40
Find the rate at which water is draining from the tank after
5 min? 3828.125 gallons/min?
10 min? 2812.5 gallons/min?
20 min? 1250 gallons / min?
40 min? 0 gal/ min?
At what time is the water flowing out the fastest?
5min-10min?
The slowest? Summarize your findings.
20min-40min?
#1, your dy/dt is correct
b) correct, to the right when v(t) > 0
but they probably want to know when that happens
look at your derivative, the denominator will always be positive, since it is something squared.
So the ± of your v(t) is only determined by the numberator which is 1 - t^2
By the "Just look at it" theorem you should be able to see that it is positive for x's between -1 and +1 , and negative for x < -1 or x> +1
c) total distance ???
find s(4) - s(0)
2. a) yes , but it asked for the max height.
the max height will be reached at 2.5 sec
so s = 80t - 16(2.5)^2 = 100
b) when s = 96
96 = -16t^2 + 80t
t^2 - 5t + 6 = 0
(t-2)(t-3) = 0
t = 2 or t = 3
so t = 2 must be on its way up and v(2) = 80 - 32(2) = 16 m/s
and t-3, on its way down v(3) = 80-3(32) = -16 m/se
3. I know nothing about Torricelli's Law, but your equation is a quadratic.
When you differentiate it, you get a linear function, and setting that equal to zero would only produce one answer, and that would give us the time when we have a minimum.
I think there is something amiss in your equation
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1. a) To find the velocity at time t, you would indeed find the derivative of x(t). Using the quotient rule, the answer would be: v(t) = (-t^2 + 1)/(t^4 + 2t^2 + 1).
b) Yes, you are correct. The particle is moving to the right when v(t) is greater than zero, and moving to the left when v(t) is less than zero.
c) To find the total distance traveled during the first 4 seconds, you would need to find the integral of |v(t)| from 0 to 4. But since v(t) is always positive or zero, the total distance traveled would simply be the integral of v(t) from 0 to 4.
2. a) To find the maximum height reached by the ball, you would find the vertex of the parabolic equation. In this case, the vertex occurs at t = -b/(2a). Using t = -80/(2*(-16)), the maximum height is reached at t = 2.5 seconds.
b) To find the velocity of the ball when it is 96 ft above the ground on its way up, you would evaluate the velocity function at that height, which means solving the equation 80t - 16t^2 = 96. Similarly, on its way down, you would solve the equation -16t^2 + 80t = 96.
3. To find the rate at which water is draining from the tank at different times, you would find the derivative of V(t) with respect to t. Based on Torricelli's Law, the rates you provided are correct.
The water is flowing out the fastest at the time when the derivative of V(t) is at its maximum value, which occurs at t = 20 minutes. The slowest rate of water flow is at t = 40 minutes when the derivative is zero, indicating that the tank is empty.
a) To find the velocity at time t, we need to take the derivative of the position function x(t). We can use the quotient rule to find the derivative:
x(t) = t/(1+t^2)
Using the quotient rule:
x'(t) = (1+t^2)(1) - t(2t) / (1+t^2)^2
Simplifying:
x'(t) = (1+t^2 - 2t^2) / (1+t^2)^2
x'(t) = (-t^2 + 1) / (1+t^2)^2
Therefore, the velocity at time t is v(t) = (-t^2 + 1) / (1+t^2)^2.
b) The particle is moving to the right when the velocity is greater than zero, and it is moving to the left when the velocity is less than zero.
For the given velocity function v(t) = (-t^2 + 1) / (1+t^2)^2, we can observe that the numerator (-t^2 + 1) is always less than or equal to zero. The denominator (1+t^2)^2 is always positive.
So, v(t) is always less than or equal to zero, except when t = 0. Therefore, the particle is always moving to the left, except at t = 0 where it is momentarily at rest.
c) To find the total distance traveled during the first 4 seconds, we need to integrate the absolute value of the velocity function from 0 to 4.
∫(|v(t)|) dt = ∫(|-t^2 + 1|) dt
Considering the limits of integration from 0 to 4:
∫(|-t^2 + 1|) dt = ∫(-(-t^2 + 1)) dt [0 to 1] + ∫(-t^2 + 1) dt [1 to 4]
Simplifying:
= ∫(t^2 - 1) dt [0 to 1] + ∫(-t^2 + 1) dt [1 to 4]
= [(t^3/3 - t) - (0)] [0 to 1] + [(-t^3/3 + t) - (t^3/3 - t)] [1 to 4]
= (1/3 - 1) - (0) + (-64/3 + 4) - (1/3 - 1)
= -2/3 + 4 + 64/3 - 3/3 + 1
= 62/3 + 1
= 65/3
Therefore, the total distance traveled during the first 4 seconds is 65/3 meters.
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For question number 2 and 3, please provide the remaining parts of the question so I can assist you better.
1) To find the velocity at time t, we need to find the derivative of x(t).
a) To find the derivative of x(t), we can use the quotient rule.
x(t) = t/(1+t^2)
We can rewrite this as x(t) = t(1+t^2)^(-1).
Now, to find the derivative, we apply the quotient rule:
x'(t) = (1+t^2)(0) - t(2t)(1+t^2)^(-2) / (1+t^2)^2
Simplifying, we get:
x'(t) = -2t^2 / (1+t^2)^2
So, the velocity at time t is v(t) = x'(t) = -2t^2 / (1+t^2)^2.
b) To determine when the particle is moving to the right or to the left, we need to analyze the sign of the velocity.
When v(t) > 0, the particle is moving to the right.
When v(t) < 0, the particle is moving to the left.
Using v(t) = -2t^2 / (1+t^2)^2, we can analyze the sign for different values of t:
For t = 0, v(t) = 0, so the particle is not moving.
For t > 0, v(t) < 0, so the particle is moving to the left.
c) To find the total distance traveled during the first 4 seconds, we need to integrate the absolute value of the velocity function from 0 to 4.
Distance traveled = ∫[0 to 4] |v(t)| dt
Using v(t) = -2t^2 / (1+t^2)^2, we can find the absolute value and integrate:
Distance traveled = ∫[0 to 4] |-2t^2 / (1+t^2)^2| dt
This integral can be a bit tricky. You can try solving it by substituting u = 1 + t^2 and then integrating by parts.
Once you find the integral, evaluate it from 0 to 4 to determine the total distance traveled during the first 4 seconds.
2) To find the answers to the questions related to the ball's height and velocity, we can use the given equation for height and apply calculus.
a) To find the maximum height reached by the ball, we need to find the vertex of the parabolic function representing its height.
The equation for the height is given as s(t) = 80t - 16t^2.
To find the maximum height, we need to find the vertex of this quadratic function. The vertex can be found using calculus or by using the formula for the x-coordinate of the vertex: -b/(2a).
In this case, a = -16 and b = 80. Plugging the values into the formula, we get:
t_vertex = -b/(2a) = -80/(2*-16) = 2.5 seconds.
So, the maximum height is reached at t = 2.5 seconds.
b) To find the velocity of the ball when it is 96 ft above the ground on its way up and on its way down, we need to find the derivative of the height function and evaluate it at the corresponding times.
Given s(t) = 80t - 16t^2, the velocity is given by the derivative:
v(t) = s'(t) = 80 - 32t.
i) When the ball is 96 ft above the ground on its way up, we have s(t) = 96. We need to solve this equation to find the time:
80t - 16t^2 = 96.
Solving this quadratic equation, we get t = 1.
To find the velocity at t = 1, we substitute t = 1 into v(t):
v(1) = 80 - 32(1) = 48 ft/s.
So, the velocity of the ball when it is 96 ft above the ground on its way up is 48 ft/s.
ii) When the ball is 96 ft above the ground on its way down, we also have s(t) = 96. Again, we solve this equation to find the time:
80t - 16t^2 = 96.
Solving this quadratic equation, we get t = 4.
To find the velocity at t = 4, we substitute t = 4 into v(t):
v(4) = 80 - 32(4) = -48 ft/s.
So, the velocity of the ball when it is 96 ft above the ground on its way down is -48 ft/s.
3) To find the rate at which water is draining from the tank at different times, we need to find the derivative of the volume function given by Torricelli's Law.
The volume function is given as V(t) = 5000(1 - t/40)^2.
To find the rate at which water is draining from the tank, we need to find the derivative of V(t) with respect to t.
V'(t) = 2(1 - t/40)(-1/40).
Simplifying this expression, we get:
V'(t) = -2/40 = -1/20.
So, the rate at which water is draining from the tank is a constant -1/20 gallons per minute.
To find the rate at specific times:
a) After 5 minutes:
V'(5) = -1/20 = -0.05 gallons/minute.
b) After 10 minutes:
V'(10) = -1/20 = -0.05 gallons/minute.
c) After 20 minutes:
V'(20) = -1/20 = -0.05 gallons/minute.
d) After 40 minutes:
V'(40) = -1/20 = -0.05 gallons/minute.
The rate is the same at all times, -0.05 gallons/minute.
To determine when the water is flowing out the fastest, we need to find the time at which the rate of change is the greatest.
From the above calculations, we see that the rate is constant at -0.05 gallons/minute throughout the 40 minutes. Therefore, the water is flowing out at the same rate throughout the entire time.