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Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,395

Differential Calculus Ex.1

1. The distance s metres moved by a particle travelling in a straight line in t seconds is given by
s=45t+11t²-t³. Find the acceleration when the particle comes to rest.

2. A particle is moving in a straight line and its distance s in time t seconds is given by
s=2t³-15t²+36t-70.
Find (i) the initial velcoity, (ii) time when the velcotiy is zero and (iii) time when acceleration is zero.

3. A stone thrown upwards has its equation of motion
s=490t-4.9t² where s is in metres and t in seconds. What is the maximum height reached by the stone?

4. The area of a circular metal plate expands when heated at a certain rate. When the radius is 2cm, it is increasing at 0.01cm/second. How fast is the area increasing?

5. What is the slope of the curve y²=x at the point (1,-2)?

6. Find the angle between the curves x²=4y and y²=4x at the point of intersection other than the origin.

7. Find the equations of tangents and normals to the following curves at the given points:-
(i) y=3x²-4x at (1,-1).
(ii) xy=c² at (ct, c/t).
(iii) y(x-2)(x-3) -x + 7 = 0 at the point where it cuts the X-axis.

8. Find the equation of the normal to the curve y=2x²-3x-2 which is parallel to the straight line x+9y-11 = 0.

9. Show that the curves y² = 4(x+1) and y² = 36(9-x) cut orthogonally.

10. Prove that the curves x = y² and xy=k cut at right angles if 8k²=1.

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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Godel

Member

Registered: 2007-03-22

Posts: 1

Re: Differential Calculus Ex.1

1. The distance s metres moved by a particle travelling in a straight line in t seconds is given by
s=45t+11t²-t³. Find the acceleration when the particle comes to rest.

To Find the Acceleration of the particle, you need to diiferentiate the given distance equation to get The velocity:
In this case,
V = ds/dt = 45 + 22t-2t^2

a = dV/dt = 22 - 4t

When the particle comes to rest, acceleration = 0, since the particle is not travelling.
and the rest is simple algebra to figure....hope this helps.

Eeyore

Member

Registered: 2007-04-16

Posts: 8

Re: Differential Calculus Ex.1

I think this depends on your definition of "at rest", I 've always thought of a particle being "at rest" when its velocity is equal to zero.  If you define "at rest" to be when the acceleration is zero, then this problem is trivial you don't need any formulas at all.
So, let's just make believe that "at rest" means when the velocity is zero.  Now now set ds/dt=o and find the time at which the paricle is "at rest", then find the acceleration at that moment.
ds/dt= 45 + 22t - 3t^2, and that equal zero when t=(11±16)/3= 9 or -5/3  , so i'm goin with the 9 cuz i'm not even sure our domain includes negative numbers, but more importantly I don't like fractions, especially fractions with denominators with factors other than 2 and 5... but that's a different story, lemma finish.  The acceleration in general is the derivative of the velocity, so a=ds²/dt²=22 - 6t, so if t = 9 then the acceleration is 22-54 = -32 and if you have a teacher that wants both of the solutions (both times when the velocity is zero), then you must include the acceleration when t = -5/3 is 22 - (-5/3)6=22 + 10 = 32 so the acceleration is the same, but in different directions.
Just my opinion, I might be right. 
Oh dear, oh my, oh no....

kelvinator44

Member

Registered: 2007-04-21

Posts: 1

Re: Differential Calculus Ex.1

In the second problem :
s=2t³-15t²+36t-70

ds/dt = 6t²-30t+36

Initial velocity is when t = 0;
Initial velocity = 36

Time when velocity is zero is :
t = 30±√(900-864)/12 = 30 ± 6/12
t=3 or 2

acceleration = dv/dt = 12t-30
Time when acceleration is zero is : 2.5