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# Gradient of the Function

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Find the gradient &#8711;g of the function g(x, y) = r^5, where r = sqrt (x^2 +y^2). Hint: introduce a new variable, u = x2 +
y2. Express g(x, y) in terms of u and use the chain rule to find dg/dx and dg/dy.

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#### Solution Summary

The gradient of a function is found. The solution is detailed and well presented.

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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 &#61636;(fg) = f &#61636; 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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