1. ## Parametric Equations!!

At time t, the position of a particle moving on a curve is given by

(a) Find the exact value of t at which the curve has a horizontal tangent.
t =

(b) Find dy/dx in terms of t.
dy/dx = (6exp(2t)-2exp(-2t))/(2exp(2t)+2exp(-2t))

(c) Find .

ok so i was able to find dy/dx, but i need help with the rest...thanks...

Originally Posted by mathlete
At time t, the position of a particle moving on a curve is given by

(a) Find the exact value of t at which the curve has a horizontal tangent.
t =

(b) Find dy/dx in terms of t.
dy/dx = (6exp(2t)-2exp(-2t))/(2exp(2t)+2exp(-2t))

(c) Find .

ok so i was able to find dy/dx, but i need help with the rest...thanks...
for c is three you have $\lim_{t \to 0}\frac{6e^{2t}-2e^{-2t}}{2e^{2t}+2e^{-2t}}$...multiply by $\frac{\frac{1}{e^{2t}}}{\frac{1}{e^{2t}}}$ and I think the answer will be apparent

3. Originally Posted by mathlete
At time t, the position of a particle moving on a curve is given by

(a) Find the exact value of t at which the curve has a horizontal tangent.
t =

(b) Find dy/dx in terms of t.
dy/dx = (6exp(2t)-2exp(-2t))/(2exp(2t)+2exp(-2t))

(c) Find .

ok so i was able to find dy/dx, but i need help with the rest...thanks...
to find horizontal tangents we need $\frac{dy}{dt}=0$

so y is

$y=3e^{2t}+e^{-2t} \iff \frac{dy}{dt}=6e^{2t}-2e^{-2t}$

setting equal to zero we get.

$0=6e^{2t}-2e^{-2t}$ Now multiply by $e^{2t}$

and we get

$0=6e^{4t}-2 \iff \frac{1}{3}=e^{4t} \iff \underbrace{\ln\left( \frac{1}{3}\right)}_{=-\ln(3)}=4t$

$-\ln(3)=4t \iff t =-\frac{1}{4}\ln(3)$