Question
At what point on the curve x = 3t^2 + 2, y=t^3-8 does the tangent line have slope 1/2?
Answers
Bosnian
x = 3 t² + 2 , y = t³ - 8
dx / dt = 3 ∙ 2 t = 6 t
dy / dt = 3 t²
( dy / dt ) / ( dx / dt ) = dy / dx = y´ = 3 t² / 6 t = t / 2
The first derivative of a function is the slope of the tangent line for any point on the function, so you must find where y' = 1 / 2
t / 2 = 1 / 2
t = 1
So :
x = 3 t² + 2 = x = 3 ∙ 1² + 2 = 3 ∙ 1 + 2 = 3 + 2 = 5
y = t³ - 8 = y = 1³ - 8 = 1 - 8 = - 7
The tangent line have slope 1 / 2 at point ( 5 , - 7 )
dx / dt = 3 ∙ 2 t = 6 t
dy / dt = 3 t²
( dy / dt ) / ( dx / dt ) = dy / dx = y´ = 3 t² / 6 t = t / 2
The first derivative of a function is the slope of the tangent line for any point on the function, so you must find where y' = 1 / 2
t / 2 = 1 / 2
t = 1
So :
x = 3 t² + 2 = x = 3 ∙ 1² + 2 = 3 ∙ 1 + 2 = 3 + 2 = 5
y = t³ - 8 = y = 1³ - 8 = 1 - 8 = - 7
The tangent line have slope 1 / 2 at point ( 5 , - 7 )
oobleck
dy/dx = dy/dt / dx/dt = 3t^2/6t = t/2
So, dy/dx = 1/2 when t=1, at (5,-7)
So, dy/dx = 1/2 when t=1, at (5,-7)
Sky
Thank you guys very much!!