Hello I'm reposting this question because right now it's at the bottom and I don't think I will get an answer.
Assume that the follwing function gives the position of an object at time t. Find the velocity indicated by setting up and evaluating a limt algebraically.
s(t) = �ãt, Find v(1).
I know that the answer is 1/2 because
s(t) = �ãt = t(1/2)
s '(t) = (1/2)t^(-1/2) = 1/(2�ãt)
s '(1) = 1/(2�ã1) = 1/2
But I can't get the answer by evaluating the limit algebraically.. can someone please show me the procedure?
let s(t) = √t
then
derivative = Limit (√(1+h) - √1)/h as h --> 0
= Lim (√(1+h) - √1)/h * (√(1+h) + 1)/(√(1+h) + 1)
= lim (1+h - 1)/(h(√(1+h) + 1))
= lim h/(h(√(1+h) + 1))
= lim 1/(√(1+h) + 1) , as h --> (in each line above)
= 1/(√1 + 1) = 1/2
Sure! I can help you with that.
To find the velocity indicated by setting up and evaluating a limit algebraically, you need to take the derivative of the position function with respect to time.
Given the position function: s(t) = √t
Step 1: Find the derivative of the position function, which represents the instantaneous velocity:
s'(t) = d/dt (√t)
Step 2: Apply the power rule for differentiation:
s'(t) = (1/2) * t^(-1/2)
Step 3: Now, we want to find the velocity at t = 1, which is represented by v(1).
To evaluate this limit algebraically, we set up the limit expression:
v(1) = lim (t->1) [s'(t)]
Step 4: Substitute the derivative function into the limit expression:
v(1) = lim (t->1) [(1/2) * t^(-1/2)]
Step 5: Evaluate the limit by substituting the value of t:
v(1) = (1/2) * (1)^(-1/2)
Now, to simplify this expression, recall that (1)^(-1/2) is equal to 1 divided by the square root of 1, which is 1/1 = 1.
v(1) = (1/2) * 1
Therefore, the velocity at t = 1 is v(1) = 1/2.
So, the answer you provided, 1/2, is indeed the correct velocity when evaluating the limit algebraically.