# For #f(t)= (sin^2t,t/pi-2)# what is the distance between #f(pi/4)# and #f(pi)#?

And

Thus the distance we want is

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To find the distance between ( f(\frac{\pi}{4}) ) and ( f(\pi) ), we can use the distance formula in two dimensions, which is the distance between two points ( (x_1, y_1) ) and ( (x_2, y_2) ) given by ( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ).

For ( f(\frac{\pi}{4}) ), ( t = \frac{\pi}{4} ), so ( f(\frac{\pi}{4}) = (\sin^2(\frac{\pi}{4}), \frac{\frac{\pi}{4}}{\pi} - 2) ).

For ( f(\pi) ), ( t = \pi ), so ( f(\pi) = (\sin^2(\pi), \frac{\pi}{\pi} - 2) ).

( \sin^2(\frac{\pi}{4}) = \sin^2(\frac{\pi}{4}) = (\frac{\sqrt{2}}{2})^2 = \frac{1}{2} ).

Now, calculate the distance using the formula:

( d = \sqrt{(\sin^2(\pi) - \sin^2(\frac{\pi}{4}))^2 + (\frac{\pi}{\pi} - 2 - (\frac{\frac{\pi}{4}}{\pi} - 2))^2} ).

Substitute the values:

( d = \sqrt{(\frac{0}{1} - \frac{1}{2})^2 + (1 - 2 - (\frac{1}{4} - 2))^2} ).

( d = \sqrt{(\frac{-1}{2})^2 + (1 - 2 - (\frac{1}{4} - 2))^2} ).

( d = \sqrt{(\frac{1}{4}) + (-1 - (\frac{-7}{4}))^2} ).

( d = \sqrt{(\frac{1}{4}) + (-1 + \frac{7}{4})^2} ).

( d = \sqrt{(\frac{1}{4}) + (\frac{3}{4})^2} ).

( d = \sqrt{(\frac{1}{4}) + (\frac{9}{16})} ).

( d = \sqrt{\frac{4}{16} + \frac{9}{16}} ).

( d = \sqrt{\frac{13}{16}} ).

( d = \frac{\sqrt{13}}{4} ).

So, the distance between ( f(\frac{\pi}{4}) ) and ( f(\pi) ) is ( \frac{\sqrt{13}}{4} ).

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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.

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.

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.

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