so pi would be substituted into
r(t) = (10 − t)i + (t 2 − 10t)j + sin tk
giving (10-2*pi) + (2*pi^2 - 10(2*pi))+sin(2*pi)
and would the equation for r'(t) be:
i + 2tj + -cosk?
so the derivative of r(t) would be:
r(t) = (10 − t)i + (t 2 − 10t)j + sin tk.
r'(t) = ti + 2tj + (-cos)tk ?
also just to back track a little the 2 * pi gets subbed into y=(10-x)^2 - 10(10-x)?
sorry i have tried.
i have so far
i.
x = 10 - t => t = 10 - x (1)
y = t^2 - 10t (2)
Substitute (1) into (2).
which gives y=(10-x)^2 - 10(10-x) which describes the particles path?
ii. and then i need to substitute 2*pi into the equation (not 100% sure which one though) to find...
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A moving particle at time t ∈ [0, 10] (seconds) has position vector in metres from the origin (0, 0, 0) given by the vector function r(t) = (10 − t)i + (t 2 − 10t)j + sin tk.
i. Describe the path of the...
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I think i need to find A(y) which equals pi(2-y)(y-1)^2
then integrate this with respect to the intervals 1 and 2 which then gives the volume?
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