IIT JEE , AIEEE Forum - Physics , Mechanics

srujana

Please explain in detail how to find the maximum and minimum of the function

h=2t-5t² , find the maximum and minimum height reached by the body.

Solve this without using the fact that at maximum height v=0

karthik2007

It is quite simple : Assuming you know differentiation, here's how you proceed :

1. Say you are given some function (which people nowadays call it by a wierd name, generally as f(x)) f(x), first differentiate it, and find f'(x).

2. Equate f'(x) to zero, and find the unknown variable. In our case, since the function is dependent on x, we will find a value for x.

3. Once you get the value of the variable, what you do is, differentiate f'(x) again, ie, find f''(x). Now, put the value of the x that you found in f''(x).

4. If f''(x) is positive, then that value of x gives the minimum value of the function. If it is negative, then it is the maximum.

let me do this sum for you :

h = 2t-5t^2.

differentiating wrt to t, we get : f'(t) = 2 - 10t.

equating it to zero, we get 10t = 2, or t = 0.2 seconds.

Differentiating again wrt to t, we get the second derivative as -10.

Since the second derivative is 0, the value of t that we got is the maximum.

Thus, t = 0.2 seconds.

Maximum height can be found by putting this value of t in the original function.

Hope it helped.

srujana

Why did u equate the first derivative to zero???????

karthik2007

See, if the function is to have a maximum or a minimum, at that point, the slope of the tangent will be zero. Thats why we equate it to 0. I know its difficult to explain without a graph. I'll try to draw a graph and explain.

neeraj_agarwal_1990

u are in class 11...so don't worry...the concept will be clear when u will do calculas....

raunak.dav

aplly the oncept of apllication of derivatives

elessar_iitkgp

You can solve it without calculus too
h = 2t-5t² = -5(t-(1/5))² + 1/5
As (t-(1/5))² is a perfect square,
(t-(1/5))²

0
-5(t-(1/5))²

0
-5(t-(1/5))² + 1/5

1/5
h

1/5
Hence maximum value of h is,
h_max = 1/5

waterdemon

Hiya Srujana met you after a long time.

I am also attaching a graph for your slope doubt.

Ok this Question can be done by Differentiation.I am giving you the rules of

I am solving the sum side by side as I am giving you the steps.

Differentiation to find Maximum and minimum value of a function :-

1)Suppose f(x) is the function given to you Then on differentiating it you can call the "Diffrentiated function" as f'(x)

So we get:

f(x) = 2t - 5t²

f'(x)= 2 - 10t

Now put the value of differentiated function = 0

f'(x)=0

2 - 10t = 0 we will get t = 1/5

2)Now differentiate the "already differentiated fuction"

f''(x) = double differentiated

f"(x)= -10

Since the value after double differentiation is < 0 we will get the minimum value of the function at t = 1/5 = 0.2sec

But as the f"'(x) = 0 we will get t = 0.2 sec as maximum.

If it would have been as derivative in terms of "t" then we would have put the value of t =1/5 in it and get its value.

And if the value is positive .minima at t =1/5

and if negative maxima at t = 1/5

the maximum height reached is

h=2t - 5t² = 2(1/5) - 5(1/25) = 2/5 - 1/5 = (1/5)

Hence the maximum height reached will be 0.2m

Hope you understood my explanation

Cheers !!!!!!!!!!!!!!!!!!!!!

karthik2007

if u are just in class XI, then chill. There's no need for you to worry about calculus now.

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