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

Sagot :

caylus

Answer:

Hello,

P=(30,6)

Step-by-step explanation:

[tex]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)\\[/tex]

The point on the curve x = 6t² + 6, y = t³ - 2 where the tangent line have slope 1/2 is (30, 6).

How to depict the point on the curve?

From the information given, x = 6t² + 6, y = t³ - 2. We'll find the first order derivative of x and y which will be:

dx/dt = 12t

dy/dt = 3t²

Therefore, 3t²/12t = t/4, t = 2.

We'll put the value of t back into the equations.

x = 6t² + 6,

x = 6(2)² + 6

x = 24 + 6 = 30

y = t³ - 2.

y = (2)³ - 2

y = 8 - 2 = 6

In conclusion, the correct options is (30, 6).

Learn more about slope on:

https://brainly.com/question/3494733

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