Vector equation question (1 Viewer)

Lith_30 · Dec 25, 2021

I need help with part a.

notme123 · Dec 25, 2021

$x=\frac{1}{2}(e^t + e^{-t})$
$2x=e^t + e^{-t}$
$2y=e^t - e^{-t}$
$(2x)^2 = e^{2t} + 2e^0 +e^{-2t}$
$(2y)^2 = e^{2t} - 2e^0 +e^{-2t}$
first eqn minus second:
$4x^2 - 4y^2 = 2-(-2) = 4$
$x^2 - y^2 =1$

fun fact:
$\sinh{x} = \frac{1}{2}(e^x - e^{-x})$
$\cosh{x} = \frac{1}{2}(e^x + e^{-x})$
and similar to the Pythagorean identity $\sin^2{x} + \cos^2{x}=1$
the identity $\sinh^2{x} - \cosh^2{x} = 1$ holds

ExtremelyBoredUser · Dec 25, 2021

@Lith_30 lmao what are these tags

Lith_30 · Dec 25, 2021

notme123 said:
$x=\frac{1}{2}(e^t + e^{-t})$
$2x=e^t + e^{-t}$
$2y=e^t - e^{-t}$
$(2x)^2 = e^{2t} + 2e^0 +e^{-2t}$
$(2y)^2 = e^{2t} - 2e^0 +e^{-2t}$
first eqn minus second:
$4x^2 - 4y^2 = 2-(-2) = 4$
$x^2 - y^2 =1$

fun fact:
$\sinh{x} = \frac{1}{2}(e^x - e^{-x})$
$\cosh{x} = \frac{1}{2}(e^x + e^{-x})$
and similar to the Pythagorean identity $\sin^2{x} + \cos^2{x}=1$
the identity $\sinh^2{x} - \cosh^2{x} = 1$ holds

Thankyou!

Lith_30 · Dec 25, 2021

ExtremelyBoredUser said:
@Lith_30 lmao what are these tags
View attachment 34516

wdym, I only select the most relevant tags

Vector equation question (1 Viewer)

Lith_30

o_o

notme123

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ExtremelyBoredUser

Bored Uni Student

Lith_30

o_o

Lith_30

o_o

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