Asked by Sky
At what point on the curve x = 3t^2 + 2, y=t^3-8 does the tangent line have slope 1/2?
Answers
Answered by
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 )
Answered by
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)
Answered by
Sky
Thank you guys very much!!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.