An object in free fall from
50 m high, so the height is
when t is h(t) = 50 – 4.9t2
.
What is the speed at
t = 2?
2. Determine the equation of the tangent line
on the curve y = x3 at the point (2.8)
1.
speed:
v(t)=h′(t)=−9.8tv(t)=h'(t)=-9.8tv(t)=h′(t)=−9.8t
v(2)=9.8⋅2=19.6v(2)=9.8\cdot 2=19.6v(2)=9.8⋅2=19.6 m/s
2.
y−y0=f′(x0)(x−x0)y-y_0=f'(x_0)(x-x_0)y−y0=f′(x0)(x−x0)
f′(x)=3x2f'(x)=3x^2f′(x)=3x2
f′(2)=3⋅22=12f'(2)=3\cdot 2^2=12f′(2)=3⋅22=12
y−8=12(x−2)y-8=12(x-2)y−8=12(x−2)
y=12x−16y=12x-16y=12x−16
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