Find the slope of f(t)= sqrt((1-sin (2t))
a) at point (0,1)
i got -1
b) at origin
i got 1
is this right thanks
f = √(1-sin(2t))
f' = 1/(2√(1-sin(2t))) * -cos(2t) * 2
= -cos(2t)/√(1-sin(2t))
at x=0, that's -1/1 = -1
you are correct
at x=0, f=1, so it makes no sense to ask about f' at the origin.
Thank you!
To find the slope of a function at a specific point, you need to take the derivative of the function and then substitute the x-coordinate of the point into the derivative. Let's calculate the derivatives and solve the problems step by step:
a) Finding the slope at the point (0,1):
To find the derivative of the given function, f(t) = √(1 - sin(2t)), we need to use the chain rule.
Let's denote u = 1 - sin(2t).
Then, applying the chain rule, we have:
f'(t) = (1/2√u) * du/dt.
To find du/dt, we differentiate u with respect to t, which gives:
du/dt = -2cos(2t).
Substituting these values back into the derivative, we get:
f'(t) = (1/2√(1 - sin(2t))) * (-2cos(2t)).
Now, let's substitute t = 0 into f'(t) to find the slope at the point (0,1):
f'(0) = (1/2√(1 - sin(2*0))) * (-2cos(2*0))
= (1/2√(1 - sin(0))) * (-2cos(0))
= (1/2√(1)) * (-2*1)
= -1.
Therefore, your answer is correct. The slope of f(t) at the point (0,1) is -1.
b) Finding the slope at the origin:
To find the slope at the origin, we need to substitute t = 0 into the derivative expression we calculated earlier: f'(t) = (1/2√(1 - sin(2t))) * (-2cos(2t)).
Substituting t = 0, we get:
f'(0) = (1/2√(1 - sin(2*0))) * (-2cos(2*0))
= (1/2√(1 - sin(0))) * (-2cos(0))
= (1/2√(1)) * (-2*1)
= -1.
The slope at the origin is -1. Therefore, your answer is incorrect. The correct slope at the origin is -1, not 1.
Please double-check your calculations again.