Question
Asked Oct 2, 2019
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F(t)=m(t)a(t)

At time t=6 seconds, the mass of an object is 36 grams and changing at a rate of −3(g/s). At this same time, the acceleration is 15(m/s^2) and changing at a rate of −8(m/s^3)

By the product rule, the force on the object is changing at the rate of...?

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Expert Answer

Step 1

The force on an object is the product of the mass m and the acceleration. In this problem, we assume both mass and acceleration are depended on time t.

F(t)m(t)*a(t).... (1)
=
We are
given the following details,
At time t
6,
dm
Mass (m) 36; Rate of change of mass
dt
da
Acceleration (a) = 15; Rate of change of acceleration
dt
-8
To Find: Rate with which the force is changing
help_outline

Image Transcriptionclose

F(t)m(t)*a(t).... (1) = We are given the following details, At time t 6, dm Mass (m) 36; Rate of change of mass dt da Acceleration (a) = 15; Rate of change of acceleration dt -8 To Find: Rate with which the force is changing

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Step 2

To determine the rate of change, we differentiate the force with respect to time (t).  

d
(F(t))
(ma)
dt
dt
Note: Product Rule:
Iffand g are both differentiable, then
d
d
[f(x) g(x)f(x) g(x) g(x)
(f(x)
dx
dx
help_outline

Image Transcriptionclose

d (F(t)) (ma) dt dt Note: Product Rule: Iffand g are both differentiable, then d d [f(x) g(x)f(x) g(x) g(x) (f(x) dx dx

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Step 3

Substitute the val...

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Tagged in

Math

Calculus

Derivative