Course Notes for Calculus III

The method of calculating directional derivatives is to take the dot product of the unit direction vector with the gradient of the function. First I calculate the gradient.

Then I calculate the unit vector for the given direction.

Then I take the dot product to produce the directional derivative.

Finally, I evaluate at the given point.

I could have first evaluated the gradient at the point and then taken the dot product with the unit direction; that order of calculations would have produced the same result.

The method of calculating directional derivatives is to take the dot product of the unit direction vector with the gradient of the function. First I calculate the gradient.

Then I calculate the unit vector for the given direction.

Then I take the dot product to produce the directional derivative.

Finally, I evaluate at the given point.

I could have first evaluated the gradient at the point and then taken the dot product with the unit direction; that order of calculations would have produced the same result.

The method of calculating directional derivatives is to take the dot product of the unit direction vector with the gradient of the function. First I calculate the gradient.

Then I calculate the unit vector for the given direction.

Then I take the dot product to produce the directional derivative.

Finally, I evaluate at the given point.

I could have first evaluated the gradient at the point and then taken the dot product with the unit direction; that order of calculations would have produced the same result.

The method of calculating directional derivatives is to take the dot product of the unit direction vector with the gradient of the function. First I calculate the gradient.

Then I calculate the unit vector for the given direction.

Then I take the dot product to produce the directional derivative.

Finally, I evaluate at the given point.

I could have first evaluated the gradient at the point and then taken the dot product with the unit direction; that order of calculations would have produced the same result.

The method of calculating directional derivatives is to take the dot product of the unit direction vector with the gradient of the function. First I calculate the gradient.

Then I calculate the unit vector for the given direction.

Then I take the dot product to produce the directional derivative.

Finally, I evaluate at the given point.

I could have first evaluated the gradient at the point and then taken the dot product with the unit direction; that order of calculations would have produced the same result.

This is a chain rule calculation. I first need the partial dervatives of the scalar field.

Then I need the derivatives of the components of the parametric curve.

Then I put these into the chain rule form and simplify.

This is a chain rule calculation. I first need the partial dervatives of the scalar field.

Then I need the derivatives of the components of the parametric curve.

Then I put these into the chain rule form and simplify. In the partial derivatives, I replace \(x\) and \(y\) with their matching components from the parametric curve.

This is a chain rule calculation. I first need the partial dervatives of the scalar field.

Then I need the derivatives of the components of the parametric curve.

Then I put these into the chain rule form and simplify. In the partial derivatives, I replace \(x\) and \(y\) with their matching components from the parametric curve.

This is a chain rule calculation. I first need the partial dervatives of the scalar field.

Then I need the derivatives of the components of the parametric curve.

Then I put these into the chain rule form and simplify. In the partial derivatives, I replace \(x\) and \(y\) with their matching components from the parametric curve.

For all of these, I will need the partial derivatives of the potential energy field.

Then I'll apply the chain rule for each curve individually and try to explain the behaviour along the curve.

1. \begin{align*} \gamma(t) \amp = (3 \cos t, 3 \sin t, 0)\\ \frac{d}{dt} \gamma_1 \amp = -3 \sin t \\ \frac{d}{dt} \gamma_2 \amp = 3 \cos t \\ \frac{d}{dt} \gamma_3 \amp = 0\\ \frac{d}{dt} f(\gamma(t)) \amp = \frac{GM}{(3\cos t)^2 + (3\sin t)^2 + 0^2)^{\frac{3}{2}}} (-3x \sin t + 3y \sin t + 0) \\ \amp = \frac{GM}{9^{\frac{3}{2}}} (-9\cos t \sin t+ 9 \sin t \cos t) = 0 \end{align*}

   This curve is a circular path around the origin. Following this curve, the distance to the origin is fixed. Potential energey only changes with distance, so it makes sense that there is no change in potential energey for this circular path.

2. \begin{align*} \gamma(t) \amp = \left( \frac{1}{t} \cos t, \frac{1}{t} \sin t, 0 \right)\\ \frac{d}{dt} \gamma_1 \amp = \frac{-1}{t^2} \cos t - \frac{1}{t} \sin t = \frac{-\cos t - t \sin t}{t^2} \\ \frac{d}{dt} \gamma_2 \amp = \frac{-1}{t^2} \sin t + \frac{1}{t} \cos t = \frac{-\sin t + t \cos t}{t^2} \\ \frac{d}{dt} \gamma_3 \amp = 0 \\ \frac{d}{dt} f(\gamma(t)) \amp = \frac{GM}{\left( \frac{1}{t} \cos t \right)^2 + \left( \frac{1}{t} \sin t \right)^2 + 0^2} \\ \amp \left( x \frac{-\cos t - t\sin t}{t^2} + y \frac{-\sin t + t \cos t}{t^2} + z(0) \right) \\ \amp = \frac{GM}{\frac{1}{t^2} + 1} \left( \frac{1}{t} \cos t \frac{-\cos t - t \sin t}{t^2} + \frac{1}{t} \sin t \frac{-\sin t + t \cos t}{t^2} \right) \\ \amp = \frac{GMt^2}{1+t^2} \frac{-\cos^t - t \sin t \cos t - \sin^2 t + t \sin t \cos t}{t^3} \\ \amp = \frac{-GM}{t(1+t^2)} \end{align*}

   This potential energy derivative is always negative, showing that the potential energy is decreasing. PE decreases as an object approach the central massive object (recall the convention that PE goes to zero at infinity and approaches \(-\infty\) as I approach the central mass). This is an inward spiral so it makes sense that potential energy is decreasing along the spiral.

3. \begin{align*} \gamma(t) \amp = (t,t^2,t^3)\\ \frac{d}{dt} \gamma_1 \amp = 1\\ \frac{d}{dt} \gamma_2 \amp = 2t\\ \frac{d}{dt} \gamma_3 \amp = 3t^2 \\ \frac{d}{dt} f(\gamma(t)) \amp = \frac{GM}{t^2 + t^4 + t^6} (x(1) + y(2t) + z(3t^2)) \\ \amp = \frac{GM}{t^2 + t^4 + t^6} (t + t^2(2t) + t^3(3t^2) = \frac{GM(1+2t^2 + 3t^4)}{t + t^3 + t^5} \end{align*}

   The derivative here is always positive, though it is getting smaller as \(t\) increases. PE increases as I move away from the central mass. This curve diverges away to \(\infty\text{,}\) so it makes sense that potential energy should decrease. As I get further and further away, the changes in PE become very small (even with the increasing speed of the curve) to it also makes sense that this derivative is getting smaller and smaller.

I can use the direct form of the equation. To use that form, I need the value of the function at the point as well as the value of its partial derivatives at the point.

Then the equation of the tangent plane is given by substituting these values in the form.

I can use the direct form of the equation. To use that form, I need the value of the function at the point as well as the value of its partial derivatives at the point.

Then the equation of the tangent plane is given by substituting these values in the form.

I can use the direct form of the equation. To use that form, I need the value of the function at the point as well as the value of its partial derivatives at the point.

Then the equation of the tangent plane is given by substituting these values in the form.

I can use the direct form of the equation. To use that form, I need the value of the function at the point as well as the value of its partial derivatives at the point.

Then the equation of the tangent plane is given by substituting these values in the form.

I can use the direct form of the equation. To use that form, I need the value of the function at the point as well as the value of its partial derivatives at the point.

Then the equation of the tangent plane is given by substituting these values in the form.

The point is not in the domain of the function. There is no tangent plane.