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For #f(t)= (sint-cost,t)# what is the distance between #f(pi/4)# and #f(pi)#?

Answer 1

Aurora Alvarado

The distance is #~~2.56#

#f(pi/4) = (sin(pi/4) - cos(pi/4), pi/4) = (0, pi/4)#

#f(pi) = (sin(pi) - cos(pi), pi) = (1, pi)#

The distance is:

#d = sqrt((1 - 0)² + (pi - pi/4)²#

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Answer 2

Samantha Adkison

To find the distance between ( f\left(\frac{\pi}{4}\right) ) and ( f(\pi) ), we first compute the values of ( f(t) ) at ( t = \frac{\pi}{4} ) and ( t = \pi ).

For ( t = \frac{\pi}{4} ), we have ( f\left(\frac{\pi}{4}\right) = (\sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4}), \frac{\pi}{4}) = (\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}, \frac{\pi}{4}) = (0, \frac{\pi}{4}) ).

For ( t = \pi ), we have ( f(\pi) = (\sin(\pi) - \cos(\pi), \pi) = (0 - (-1), \pi) = (1, \pi) ).

Now, we calculate the distance between these two points using the distance formula:

[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

Substituting the coordinates:

[ d = \sqrt{(1 - 0)^2 + (\pi - \frac{\pi}{4})^2} = \sqrt{1 + \left(\frac{3\pi}{4}\right)^2} = \sqrt{1 + \frac{9\pi^2}{16}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.