Calc II - PARAMETRIC EQUATIONS
Find an equation of the tangent to the curve at the point by first eliminating the parameter.

Question

Calc II - PARAMETRIC EQUATIONS
Find an equation of the tangent to the curve at the point by first eliminating the parameter.

anonymous · Answer

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rogue · Answer

$$x = e^t, y = (t - 6)^2$$$$x = e^t \rightarrow t = \ln x \rightarrow y = (\ln x - 6)^2$$$$y' = \frac {2(\ln x - 6)}{x}$$$$y' (1) = -12$$$$Y_{tangent} = 36 - 12(x -1) = 37 - 12x$$

rogue · Answer

Oops, I didn't distribute the 12 to the 1, it should be Y = 48 - 12x

anonymous · Answer

Thanks rogue!

rogue · Answer

yw :)

campbell_st · Answer

if

$$x = e^x$$

then 

$$\ln _{e} x = t$$

sub into the y equation

$$y = ( \ln (x) - 6)^2$$

differentiate

$$y' = 2/x \times (\ln(x) - 6)$$

substitute x = 1 to find the gradient

$$m = 2/1 \times (\ln(1) - 6) = -12$$

so the point is (1, 36) and gradient is -12 

use the point slope formula to find the equation of the tangent

$$y - 36 = -12(x -1)$$

y = -12x + 48

anonymous · Answer

Wow, you went the full 5 yards on this one, thanks for explaining it!
Much appreciated!