Application of calculas questions? (1 Viewer)

[RHINO7]

I can not work out these questions---

i)The rate of change in velocity over time is given by dy/dt=4t+t^2-t^3. If the intial velocity is 2cms^-1, find the velocity after 15s.

ii)The number of people in a certain city is given by P=200000e^0.2t, where t is time in years. Find the rate at which the population of the city will be growing after 5 years.



Is this topic the hardest?

 

[Forbidden.]

i)

dy/dt = 4t + t² - t³ (Find the velocity)
y = ∫ 4t + t² - t³ dt
y = 2t² + t³/3 - t⁴/4 + C

When t = 0, y = 2

2 = 2(0)² + (0)³/3 - (0)⁴/4 + C (Find the constant)
C = 2
y = 2t² + t³/3 - t⁴/4 + 2

When t = 15

y = 2(15)² + (15)³/3 - (15)⁴/4 + 2 (After 15 seconds what's the velocity ?)
y = - 11,081.25 ms^-1


RATE OF CHANGE means the DERIVATIVE of a function.
E.g to find the FUNCTION you INTEGRATE the RATE OF CHANGE OF THE FUNCION.
E.g to find the VELOCITY you INTEGRATE the RATE OF CHANGE OF THE VELOCITY.

Conversely

To find the RATE OF CHANGE OF THE VELOCITY you DIFFERENTIATE the VELOCITY.
To find the RATE OF CHANGE OF THE FUNCTION you DIFFERENTIATE the FUNCTION.



ii)

If this isn't just substituting t = 5 in P=200000e^0.2t then,

dP/dt = 40,000e^0.2t (Find the rate of change)
When t = 5
dP/dt = 40,000e^{0.2 x 0.5}
dP/dt = 40,000e^0.1
dP/dt = 44,206.83672 ...
dP/dt = 44,207 (to whole number)

RATE means derivative.

.: The rate of population growth after 5 years is 44,207 people per year
(Which means, during the 5th year, each new year, approximately 44,207 people will be introduced into the population)


Is this topic the hardest?

It's very popular, often popular questions are added deliberately because they are known for poor responses to them.
Beware of these questions in your trials and HSC, the Physicists may have the upper hand in this topic.

 

[RHINO7]

THANKS however the answer for ii) is 108 731.

 

[jannny]

Question 1 was kinda tricky cos normally if it is dy/dt, it is often referred as the Velocity already.

 

[jannny]

i)



If this isn't just substituting t = 5 in P=200000e^0.2t then,

dP/dt = 40,000e^0.2t (Find the rate of change)
When t = 5
dP/dt = 40,000e^{0.2 x 0.5}
dP/dt = 40,000e^0.1
dP/dt = 44,206.83672 ...
dP/dt = 44,207 (to whole number)

should have been
dp/dt = 40,000 x e^ (0.2 x 5)
dp/dt = 40,000 x e1
= 108,731.2731
=108, 731 (nearest person as can't have a 0.2 person)

 

[Forbidden.]

should have been
dp/dt = 40,000 x e^ (0.2 x 5)
dp/dt = 40,000 x e1
= 108,731.2731
=108, 731 (nearest person as can't have a 0.2 person)

olol .. That is my greatest weakness in Mathematics and is of great concern, stupid mistakes. Thanks for the correction.