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Force as the Gradient of Potential Energy

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1) Find the partial derivatives with respect to x, y, and z of the following functions: (a) f(x, y, z) = ax2 + bxy + cy2, (b) g(x, y, z) = sin(axyz2), (c) h(x, y, z) = aexy/z^2, where a, b, and c are constants.

2) Find the partial derivatives with respect to x, y, and z of the following functions: (a) f(x, y, z) = ay2 + 2byz + cz2, (b) g(x, y, z) = cos(axy2z3), (c) h(x, y, z) = ar, where a, b, and c are constants and r = sqrt(x2 + y2 + z2).

3) Calculate the gradient of the following functions, f(x, y, z): (a) f = x2 + z3, (b) f = ky, where k is a constant, (c) f = r sqrt(x2 + y2 + z2), (d) f = 1/r.

4) Calculate the gradient of the following functions, f(x, y, z): (a) f = ln(r), (b) f = r2, (c) f = g(r), where r = sqrt(x2 + y2 + z2) and g(r) is some unspecified function of r.

5) Prove that if f(r) and g(r) are any two scalar functions of r, then (fg) = f  g + g f.

6) If a particle's potential energy is U(r) = k(x2 + y2 + z2), where k is a constant, what is the force on the particle?

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Solution Summary

Word document find partial derivatives, calculates gradients and proves that two functions are scalar.

Solution provided by:
Changping Wang, MA
Education
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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