Question

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 4 ln(t), y = t2 2, (4, 3)

Answer 1

I assume the curve has parametric equations

`\begin{cases} x(t) = 4 \ln(t) \\ y(t) = t^2 + 2 \end{cases}`

Eliminating the parameter:

Solve for
`t`.

`x = 4\ln(t) \implies \ln(t) = \frac x4 \implies t = e^(x/4)`

Substitute this into
`y(t)`.

`y = \left(e^(x/4)\right)^2 + 2 = e^(x/2) + 2`

Compute
`(dy)/(dx)` and evaluate it at (4, 3) (that is, with
`x=4`) to find the slope of the tangent line at that point.

`(dy)/(dx) = \frac12 e^(x/2) \implies (dy)/(dx)\bigg|_(x=4) = \frac12 e^(4/2) = \frac{e^2}2`

Use the point-slope formula to get the equation of the line.

`y - 3 = \frac{e^2}2 (x - 4) \implies \boxed{y = \frac{e^2}2 x + 3 - 2e^2}`

Without eliminating the parameter:

Use the chain rule to compute
`(dy)/(dx)`.

`(dy)/(dx) = (dy)/(dt)\cdot (dt)/(dx) = ((dy)/(dt))/((dx)/(dt))`

`(dy)/(dx) = (2t)/(\frac4t) = \frac{t^2}2`

When
`x=4`, we have

`4=4\ln(t) \implies \ln(t)=1 \implies t = e^1 = e`

and so at (4, 3), the slope of the tangent is

`(dy)/(dx)\bigg|_(t=e) = \frac{e^2}2`

Then using the point-slope formula, the tangent line's equation is again

`y - 3 = \frac{e^2}2 (x - 4)`