By first principle find the derivative of y=√t at the point (1,1). Discuss the behaviour of the function at point (0,0)

Answer 1

To find the derivative of the function y = √t by first principles, we need to evaluate the limit of the difference quotient as h approaches 0. The difference quotient is given by:

f'(t) = lim(h→0) [(√(t+h) - √t)/h]

Let's simplify this expression by rationalizing the numerator:

f'(t) = lim(h→0) [(√(t+h) - √t)/h] * [(√(t+h) + √t)/(√(t+h) + √t)]
= lim(h→0) [(t+h - t)/(h(√(t+h) + √t))]
= lim(h→0) [h/(h(√(t+h) + √t))]
= lim(h→0) [1/(√(t+h) + √t)]

Now we can evaluate the limit as h approaches 0:

f'(t) = 1/(√(t+0) + √t)
= 1/(2√t)

So the derivative of y = √t is 1/(2√t).

Now let's discuss the behavior of the function at the point (0,0). Since the point (0,0) lies on the y-axis, the function is not defined at t = 0. This is because the square root of 0 is undefined. Therefore, the function does not have a behavior at the point (0,0).

Answer 2

surely you know that √0 = 0, so it is well defined.
The only problem is that the slope is undefined, since the tangent there is vertical.

Answer 3

You are correct, and I apologize for the error in my previous response. The point (0,0) is indeed well defined in the context of √t, and its value is 0. However, the behavior of the function at that point is different.

To discuss the behavior of the function at the point (0,0), we can consider the limit as t approaches 0. Let's find this limit:

lim(t→0) √t = √0 = 0

Therefore, as t approaches 0, the value of √t also approaches 0. This means that the function approaches the y-axis as t approaches 0. However, since the derivative at t = 0 is undefined (as you correctly mentioned), the function does not have a well-defined slope at the point (0,0). The tangent to the curve at that point is vertical, indicating a sharp change in direction.