Differential calculus, physics problem

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SUMMARY

The discussion centers on finding the velocity function for the motion of a spring modeled by the equation s(t) = 5e^−1.9t sin(2πt). The user initially derived the velocity function using the product and chain rules but faced issues with grading by an automated system. It was later clarified that the assignment required a step-by-step differentiation of the position function, making the derived velocity function unnecessary for the initial parts of the problem.

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  • Understanding of differential calculus, specifically the product and chain rules.
  • Familiarity with exponential and trigonometric functions.
  • Basic knowledge of motion equations in physics.
  • Experience with automated grading systems for mathematical functions.
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  • Review the product rule and chain rule in calculus for differentiation.
  • Practice deriving velocity and acceleration functions from position functions.
  • Explore the implications of damping forces in physics, particularly in spring motion.
  • Investigate common pitfalls in automated grading systems for calculus problems.
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Students studying calculus and physics, particularly those dealing with motion equations and differentiation, as well as educators looking to understand common challenges in automated grading systems.

dlp248
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Homework Statement



The motion of a spring that is subject to a frictional force or a damping force (such as a shock absorber in a car) is often modeled by the product of an exponential function and a sine or cosine function. Suppose the equation of motion of a point on such a spring is
s(t) = 5e^−1.9t sin 2πt
where s is measured in centimeters and t in seconds. Find the velocity after t seconds. Graph both the position and velocity functions for 0 ≤ t ≤ 2.

Homework Equations



Chain rule: [f(g(x)]' = f'(g(x))g'(x)
Product rule: [f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)

The Attempt at a Solution



I am stuck finding the velocity function. I believe that I have the correct derivative, however, I am being marked incorrect.

This is what I have done:

v(t) = (5e^-1.9t)(-1.9)sin(2πt)+(5e^-1.9t)cos(2πt)(2π)
v(t) = 5e^-1.9t[-1.9sin(2πt)+2π(cos(2πt)]

I haven't attempted the rest of the problem because I need the velocity function first.
 
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dlp248 said:

Homework Statement



The motion of a spring that is subject to a frictional force or a damping force (such as a shock absorber in a car) is often modeled by the product of an exponential function and a sine or cosine function. Suppose the equation of motion of a point on such a spring is
s(t) = 5e^−1.9t sin 2πt
where s is measured in centimeters and t in seconds. Find the velocity after t seconds. Graph both the position and velocity functions for 0 ≤ t ≤ 2.

Homework Equations



Chain rule: [f(g(x)]' = f'(g(x))g'(x)
Product rule: [f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)

The Attempt at a Solution



I am stuck finding the velocity function. I believe that I have the correct derivative, however, I am being marked incorrect.

This is what I have done:

v(t) = (5e^-1.9t)(-1.9)sin(2πt)+(5e^-1.9t)cos(2πt)(2π)
v(t) = 5e^-1.9t[-1.9sin(2πt)+2π(cos(2πt)]

I haven't attempted the rest of the problem because I need the velocity function first.
It's not clear why your velocity function is being marked incorrect. Is it due to your work being examined by some computer response system?

Perhaps submitting a different version of the same function might help. I would try moving the constant 5 inside the [] brackets, leaving the exponential outside by itself.

I don't have much experience with such systems, but I can see why my late friend called computers "The Devil's Machine".
 
Thanks for getting back to me!

I was entering this function into a computer for grading. I ended up contacting my professor and she was able to point out to me that the question was asking for a step by step differentiation of the position function. So the velocity function that I found ended up not being needed until part three of the question. Why they do this to us I have no idea! It is frustrating beyond belief! Thank you again!
 

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