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Darina [25.2K]
3 years ago
10

The parametric equations of a curve are x=e^-t*cost y=e^-t*sint show that dy/dx=tan(t-pi/4)

Mathematics
1 answer:
Paraphin [41]3 years ago
3 0
\bold {\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{ (\frac{dx}{dt})}}

Using product rule: f'g + fg'
\frac{dy}{dt} = -e^{-t} sin (t) + e^{-t} cos (t) \\  \\ \frac{dx}{dt} = -e^{-t} cos (t) - e^{-t} sin (t)

Factor out -e^{-t}

\bold {\frac{dy}{dx} = \frac{sin (t) - cos(t)}{sin (t) + cos (t)}}

Next, we have to use trig identities to equate this with tan(t - pi/4).
There are a lot of steps here, i think its easiest to start at either end and meet in the middle. I will show that:
tan(t - \frac{\pi}{4}) = tan(2t) - sec(2t) = \frac{sin (t) - cos(t)}{sin(t)+cos(t)}

Starting on left side, use angle sum identity for tan. Note tan(pi/4) = 1.
tan(t - \frac{\pi}{4}) = \frac{tan(t) - 1}{1+tan(t)} \\  \\  = \frac{tan(t) - 1}{1+tan(t)}*\frac{1-tan(t)}{1-tan(t)} \\  \\ =\frac{2 tan(t) - (1+tan^2(t))}{1-tan^2(t)} \\  \\ = \frac{2 tan(t) - sec^2 (t))}{1-tan^2(t)} \\  \\ = tan(2t) - \frac{sec^2 (t))}{1-tan^2(t)} \\  \\ =tan(2t) - \frac{1}{cos^2(t) - sin^2(t)} \\  \\ = tan(2t) - sec(2t)

Now starting on right side:
\frac{sin(t) - cos(t)}{sin(t) + cos(t)} \\  \\ =\frac{sin(t) - cos(t)}{sin(t) + cos(t)} * \frac{sin(t) - cos(t)}{sin(t)-cos(t)} \\  \\ = \frac{sin^2(t) + cos^2(t) -2sin(t)cos(t)}{sin^2(t) -cos^2(t)} \\  \\ = \frac{1-sin(2t)}{-cos(2t)} \\  \\ =tan(2t) - sec(2t)

Therefore proving that 
\bold{\frac{dy}{dx} = \frac{sin(t) - cos(t)}{sin(t) + cos(t)} = tan(t - \frac{\pi}{4})}
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