Question

At what point on the curve x = 6t2 + 6, y = t3 − 2 does the tangent line have slope 1 /2 ?

Answer 1

Answer:

Hello,

P=(30,6)

Step-by-step explanation:

$x=6t^2+6\\y=t^3-2\\\\\dfrac{dx}{dt}= 12t\\\dfrac{dy}{dt}= 3t^2\\\\\dfrac{dy}{dx} =\dfrac{\dfrac{dy}{dt} }{\dfrac{dx}{dt} } =\dfrac{3t^2}{12t} =\dfrac{t}{4} \\\\\dfrac{t}{4} =\dfrac{1}{2} \Longrightarrow t=2\\\\\\x=6t^2+6=6*2^2+6=30\\\\y=t^3-2=2^2-2=8-2=6\\\\\\Tangence\ point=(30,6)$