s = t^3-9t^2+3t+1
v = 3t^2-18t+3 = 3(t^2-6t+1)
so, solve for t when v=-24
a = 6t-18
then plug in that t to find v and a.
what is your problem with minus signs (-) ? over- and under-score really are ugly!
x^2 cosy - siny = 0
Surely you can plug in the point to verify it fits!
for dy/dx, just use the chain and product rules:
2x cosy - x^2 siny y' - cosy y' = 0
now just collect terms and solve for y':
y' = (2x cosy)/(x^2 siny + cosy)
or, if cosy≠0,
2x/(x^2 tany + 1)
1 A particle is moving in a stright line with the position at any time t is given by s= t³–9t²+3t+1,where s in meters and t in second. Find its position and acceleration when velocity is ¯24m/s
2. Given that x²cos y_sin y=0,(0,π).
A. Verify that the given points on the curve.
B.use implicit differention to find the slope of the above curve at the given point.
C.find the equation of tangent and normal to the curve at that.
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