find an equation of the tangent to the curve at the point corresponding to the given value of the…

find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.\n$x = e^{\\sqrt{t}}$, $y = t - 2\\ln t$; $t = 1$\nshow all steps clearly. enter your final answer.\nequation of the tangent to the curve________________________

find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.\n$x = e^{\\sqrt{t}}$, $y = t - 2\\ln t$; $t = 1$\nshow all steps clearly. enter your final answer.\nequation of the tangent to the curve________________________

Answer

Explanation:

Step1: Find (x) and (y) at (t = 1)

When (t=1), (x = e^{\sqrt{1}}=e), (y=1 - 2\ln1=1). So the point ((x,y)=(e,1))

Step2: Find (\frac{dx}{dt}) and (\frac{dy}{dt})

  • For (x = e^{\sqrt{t}}), using the chain rule (\frac{dx}{dt}=e^{\sqrt{t}}\cdot\frac{1}{2\sqrt{t}})
  • For (y=t - 2\ln t), (\frac{dy}{dt}=1-\frac{2}{t})

Step3: Calculate (\frac{dy}{dx})

By the formula (\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}), substitute (\frac{dx}{dt}=e^{\sqrt{t}}\cdot\frac{1}{2\sqrt{t}}) and (\frac{dy}{dt}=1 - \frac{2}{t}) (\frac{dy}{dx}=\frac{1-\frac{2}{t}}{e^{\sqrt{t}}\cdot\frac{1}{2\sqrt{t}}}) When (t = 1), (\frac{dy}{dx}=\frac{1-2}{e\cdot\frac{1}{2}}=\frac{- 1}{\frac{e}{2}}=-\frac{2}{e})

Step4: Use the point - slope form (y - y_0=m(x - x_0))

Here (x_0 = e), (y_0 = 1), (m=-\frac{2}{e}) (y - 1=-\frac{2}{e}(x - e)) Expand: (y-1=-\frac{2}{e}x + 2) (y=-\frac{2}{e}x+3)

Answer:

(y =-\frac{2}{e}x + 3)