For #f(t)= (t/sqrt(t+1),t^2-t)# what is the distance between #f(0)# and #f(2)#?

Question

Ezra Adams · Accepted Answer

#(4sqrt(3))/3# #f(0) = (0/(sqrt(0+1)), 0^2-0)# #= (0, 0)# #f(2) = (2/(sqrt(2+1)), 2^2-2)# #= (2/sqrt(3), 2)# Let's use the distance formula. In case you don't know what it is, here it is: The distance #d# between #(a, b)# and #(c, d)# is: #d = sqrt((d-b)^2 + (c-a)^2)#. We can apply this to our problem. Using the distance formula, #d = sqrt((2-0)^2 + (2/sqrt(3)-0)^2)# #= sqrt((2^2) + (2/sqrt(3))^2)# #=sqrt(4 + 4/3)# #=sqrt(12/3 + 4/3)# #=sqrt(16/3)# #=4/sqrt(3)# #=(4sqrt(3))/3# (if you want to rationalize the denominator)

Scarlett Allison · Answer

undefined To find the distance between $ f(0) $ and $ f(2) $ for the function $ f(t) = \left(\frac{t}{\sqrt{t+1}}, t^2 - t\right) $, we'll calculate the distance using the distance formula in two-dimensional space:

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

where $ f(0) = (x_1, y_1) $ and $ f(2) = (x_2, y_2) $.

First, let's find $ f(0) $ and $ f(2) $:

For $ f(0) $:
$$ t = 0 $$
$$ x_1 = \frac{0}{\sqrt{0+1}} = 0 $$
$$ y_1 = 0^2 - 0 = 0 $$
So, $ f(0) = (0, 0) $.

For $ f(2) $:
$$ t = 2 $$
$$ x_2 = \frac{2}{\sqrt{2+1}} = \frac{2}{\sqrt{3}} $$
$$ y_2 = 2^2 - 2 = 2 $$
So, $ f(2) = \left(\frac{2}{\sqrt{3}}, 2\right) $.

Now, we'll use the distance formula:

$$ d = \sqrt{\left(\frac{2}{\sqrt{3}} - 0\right)^2 + \left(2 - 0\right)^2} $$

$$ d = \sqrt{\frac{4}{3} + 4} $$

$$ d = \sqrt{\frac{4}{3} + \frac{12}{3}} $$

$$ d = \sqrt{\frac{16}{3}} $$

$$ d = \frac{4}{\sqrt{3}} $$

Thus, the distance between $ f(0) $ and $ f(2) $ is $ \frac{4}{\sqrt{3}} $.