A curve has parametric equations: x = e^t

Question

A curve has parametric equations: x = e^t - 2e^-t and y = 3e^2t + 1.  Find the equation to the curve at at the point for which t = 0.

nick1337 · Accepted Answer

We are given the parametric equations of a curve:x=et-2e-ty=3e2t+1We need to find the equation to the curve at the point where t = 0.To find the equation to the curve, we need to use the formula:dydx=dydtdxdtFirst, let&#039;s find&amp;nbsp;dxdt:dxdt=et+2e-tNow, let&#039;s find&amp;nbsp;dydt:dydt=6e2tUsing these values, we can find&amp;nbsp;dydx:dydx=dydtdxdtdydx=6e2tet+2e-tWhen&amp;nbsp;t=0, we can substitute the value in the expression for&amp;nbsp;dydx&amp;nbsp;to get the slope of the curve at that point:dydx=6e0e0+2e0=2Therefore, the slope of the curve at the point where t = 0 is 2. To find the equation to the curve at this point, we need to find the y-intercept. We can use the equation of the curve to find the value of y when x = 0:x=et-2e-t0=et-2e-tet=2e-te2t=2t=ln(2)2Now, we can substitute&amp;nbsp;t=ln(2)2&amp;nbsp;into the equation for y to find the value of y when x = 0:y=3e2t+1y=3eln(2)+1y=3(2)+1y=7Therefore, the y-intercept of the curve is 7. Now, we can use the point-slope form of a linear equation to write the equation of the curve at the point where t = 0:y-7=2(x-0)y-7=2xy=2x+7Therefore, the equation of the curve at the point where&amp;nbsp;t=0&amp;nbsp;is&amp;nbsp;y=2x+7.

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