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Question

For

f (t) = (\ln t - e^{t}, \frac{t^{2}}{e^{t}})

what is the distance between f(2) and f(4)?

Answer 1

Step 1
The distance between two points

P 1 (x 1, y 1)

and

P 2 (x 2, y 2)

is given as

\sqrt{{(x 1 - x 2)}^{2} + {(y 1 - y 2)}^{2}}

Step 2
with

f (t) = (\ln (t) - e^{t}, \frac{t^{2}}{e^{t}})

we get

f (2) = (\ln (2) - e^{2}, \frac{4}{e^{2}})

f (4) = (\ln (4) - e^{4}, \frac{16}{e^{4}})

so the distance is given by

\sqrt{{(e^{4} - e^{2} - \ln (2))}^{2} + {(\frac{4}{e^{2}} - \frac{16}{e^{4}})}^{2}}

Answers (1)