thaddom
21.07.2020 •
Mathematics
Find
dy/dx and d2y/dx2,
and find the slope and concavity (if possible) at the given value of the parameter. (If an answer does not exist, enter DNE.)
Parametric Equations Point
x = 5t, y = 6t − 1
t = 2
Solved
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Ответ:
Step-by-step explanation:
Given the parametric equation points x = 5t, y = 6t − 1 when t = 2
From x = 5t, t = x/5. Substituting t = x/5 into the second equation y = 6t − 1 we will have;
y = 6(x/5) - 1
y = 6/5 x - 1
The derivative of y with respect to x i.e dy/dx = 6/5 - 0. (Note that differential of any constant is zero).
dy/dx = 6/5
d²y/dx² = d/dx(dy/dx)
d²y/dx² = d/dx(6/5)
Since 6/5 is a constant, the derivative of 6/5 with respect to x will be zero.
d²y/dx² = 0.
Since the first derivative and the second derivative are both constant then, the slope m at the given parameter will be 6/5.
m = dy/dx = 6/5
The concavity is the value of the second derivative at the given value of the parameter.
The concavity d²y/dx² = 0.
Ответ:
umm why would u do that and sure I will help u out but geez
Step-by-step explanation: