Answer:
Step-by-step explanation:
Given that:
[tex]x = 5 + In (t)[/tex]
[tex]y = t^2+2[/tex]
At point (5,3)
To find an equation of the tangent to the curve at the given point,
By without eliminating the parameter
[tex]\dfrac{dx}{dt}= \dfrac{1}{t}[/tex]
[tex]\dfrac{dy}{dt}= 2t[/tex]
[tex]\dfrac{dy}{dx}= \dfrac{ \dfrac{dy}{dt} }{\dfrac{dx}{dt} }[/tex]
[tex]\dfrac{dy}{dx}= \dfrac{ 2t }{\dfrac{1}{t} }[/tex]
[tex]\dfrac{dy}{dx}= 2t^2[/tex]
[tex]\dfrac{dy}{dx}_{ (5,3)}= 2t^2_{ (5,3)}[/tex]
t² + 5 = 4
t² = 4 - 5
t² = - 1
Then;
[tex]\dfrac{dy}{dx}_{ (5,3)}= -2[/tex]
The equation of the tangent is:
[tex]y -y_1 = m(x-x_1)[/tex]
[tex](y-3 )= -2(x - 5)[/tex]
y - 3 = -2x +10
y = -2x + 7
y = 2x - 7
By eliminating the parameter
x = 5 + In(t)
In(t) = 5 - x
[tex]t =e^{x-5}[/tex]
[tex]y = (e^{x-5})^2+5[/tex][tex]y = (e^{2x-10})+5[/tex]
[tex]\dfrac{dy}{dx} = 2e^{2x-10}[/tex]
[tex]\dfrac{dy}{dx}_{(5,3)} = 2e^{10-10}[/tex]
[tex]\dfrac{dy}{dx}_{(5,3)} = 2[/tex]
The equation of the tangent is:
[tex]y -y_1 = m(x-x_1)[/tex]
[tex](y-3 )= -2(x - 5)[/tex]
y - 3 = -2x +10
y = -2x + 7
y = 2x - 7
