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 = 5 + ln(t), y = t2 + 2, (5, 3)

Answer 1

Answer:

Step-by-step explanation:

Given that:

$x = 5 + In (t)$

$y = t^2+2$

At point (5,3)

To find an equation of the tangent to the curve at the given point,

By without eliminating the parameter

$\dfrac{dx}{dt}= \dfrac{1}{t}$

$\dfrac{dy}{dt}= 2t$

$\dfrac{dy}{dx}= \dfrac{ \dfrac{dy}{dt} }{\dfrac{dx}{dt} }$

$\dfrac{dy}{dx}= \dfrac{ 2t }{\dfrac{1}{t} }$

$\dfrac{dy}{dx}= 2t^2$

$\dfrac{dy}{dx}_{ (5,3)}= 2t^2_{ (5,3)}$

t²  + 5 = 4

t² = 4 - 5

t² = - 1

Then;

$\dfrac{dy}{dx}_{ (5,3)}= -2$

The equation of the tangent  is:

$y -y_1 = m(x-x_1)$

$(y-3 )= -2(x - 5)$

y - 3 = -2x +10

y = -2x + 7

y = 2x - 7

By eliminating the parameter

x = 5 + In(t)

In(t)  = 5 - x

$t =e^{x-5}$

$y = (e^{x-5})^2+5$$y = (e^{2x-10})+5$

$\dfrac{dy}{dx} = 2e^{2x-10}$

$\dfrac{dy}{dx}_{(5,3)} = 2e^{10-10}$

$\dfrac{dy}{dx}_{(5,3)} = 2$

The equation of the tangent  is:

$y -y_1 = m(x-x_1)$

$(y-3 )= -2(x - 5)$

y - 3 = -2x +10

y = -2x + 7

y = 2x - 7