At least two people had got these answers and they were incorrect. Can someone p

Question

At least two people had got these answers and they were incorrect. Can someone please help

Consider the parametric curve: x = 64 t^2 y = t^3 - 9t At which t value(s) is the tangent to this curve vertical? (Note: If there is more than one write them as a comma separated list) t =___Write the point(s) at which the tangent is vertical.(Note: If there is more than one write them as a comma separated list, e.g. (1, 2),(3,4)) At which t value(s) is the tangent to this curve horizontal? (Note: If there is more than one write them as a comma separated list). t =__Write the point(s) at which the tangent is horizontal.(Note: If there is more than one write them as a comma separated list, e.g. (1, 2),(3,4)) Use a graphing device to graph the curve and check your work.

Explanation / Answer

Only last part is wrong. Lets solve it
From part iii, t values are 3/sqrt(3) and -3/sqrt(3)

for t = 3/sqrt(3)
x = 64 - (3/sqrt(3))^2
= 64 - 9/3
= 61
y = (3/sqrt(3))^3 - (3/sqrt(3))^2
= 27/3sqrt(3) - 9/3
= 9/sqrt(3) - 3

for t = -3/sqrt(3)
x = 64 - (-3/sqrt(3))^2
= 64 - 9/3
= 61
y = (-3/sqrt(3))^3 - (-3/sqrt(3))^2
= -27/3sqrt(3) - 9/3
= -9/sqrt(3) - 3

So your answer should be:
(61 ,9/sqrt(3) - 3 )
(61, -9/sqrt(3) - 3)

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