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Course Notes for Multivariable Calculus

Remkes Kooistra

Section 4.6 Week 4 Activity

Subsection 4.6.1 Directional Derivatives

Activity 4.6.1.

Calculate the directional derivative of \(f(x,y) = 3x + 5y - 2xy + x^2 - y^2 \) at the point \(p = (0,-2) \) in the direction \((-1,-1) \text{.}\) (Don’t forget to make the direction a unit vector.)
Solution.
First I need to make the direction a unit vector. The length of the direction is
\begin{equation*} |v| = \sqrt{ (-1)^2 + (-1)^2 } = \sqrt{2} \end{equation*}
so the unit vector is
\begin{equation*} u = \frac{v}{|v|} = \left( \frac{-1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right) \text{.} \end{equation*}
Then I calculate the gradient of \(f\text{,}\) taking the partials in each variable.
\begin{equation*} \nabla f = \left( (3-2y + 2x, 5 - 2x - 2y \right) \end{equation*}
Finally, the directional derivative is the dot product of gradient and the direction.
\begin{equation*} D_u f = \frac{1}{\sqrt{2}} (-3 + 2y - 2x -5 + 2x + 2y ) = \frac{-8 + 4y}{\sqrt{2}} \end{equation*}
Finally I eavluate this at the point.
\begin{equation*} D_u f (p) = \frac{-8 - 8}{\sqrt{2}} = \frac{-16}{\sqrt{2}} \end{equation*}

Activity 4.6.2.

Calculate the directional derivative of \(f(x,y) = \cos (x - y) \) at the point \(p = \left( \frac{\pi}{2}, \frac{\pi}{2} \right) \) in the direction \((2,2) \text{.}\) (Don’t forget to make the direction a unit vector.)
Solution.
First I need to make the direction a unit vector. The length of the direction is
\begin{equation*} |v| = \sqrt{ 2^2 + 1^1 } = \sqrt{5} \end{equation*}
so the unit vector is
\begin{equation*} u = \frac{v}{|v|} = \left( \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right) \text{.} \end{equation*}
Then I calculate the gradient of \(f\text{,}\) taking the partials in each variable.
\begin{equation*} \nabla f = \left( -\sin (x - y), \sin (x - y) \right) \end{equation*}
Finally, the directional derivative is the dot product of gradient and the direction.
\begin{equation*} D_u f = \frac{1}{\sqrt{5}} (-2 \sin (x - y) + \sin (x- y)) = \frac{-\sin (x - y)}{\sqrt{5}} \end{equation*}
Finally I eavluate this at the point.
\begin{equation*} D_u f (p) = \frac{-\sin (0)}{\sqrt{5}} = 0 \end{equation*}

Activity 4.6.3.

Calculate the directional derivative of \(f(x,y) = xy - yz + yz^2 - x^3 \) at the point \(p = (3,1,-2) \) in the direction \((1,3,1) \text{.}\)
Solution.
First I need to make the direction a unit vector. The length of the direction is
\begin{equation*} |v| = \sqrt{1^2 + 3^2 + 1^2} = \sqrt{11} \end{equation*}
so the unit vector is
\begin{equation*} u = \frac{v}{|v|} = \left( \frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{1}{\sqrt{11}} \right) \text{.} \end{equation*}
Then I calculate the gradient of \(f\text{,}\) taking the partials in each variable.
\begin{equation*} \nabla f = \left(y - 3x^2, x - z + z^2, -y + 2yz \right) \end{equation*}
Finally, the directional derivative is the dot product of gradient and the direction.
\begin{equation*} D_u f = \frac{1}{\sqrt{11}} (y - 3x^2 + 3x - 3z + 3x^2 - y + 2yz) \end{equation*}
Finally I eavluate this at the point.
\begin{equation*} D_u f (p) = \frac{1}{\sqrt{11}} (1 - 27 + 9 + 6 + 12 - 1 - 4) = \frac{-4}{\sqrt{11}} \end{equation*}

Activity 4.6.4.

Calculate the directional derivative of \(f(x,y) = e^{xyz}\) at the point \(p = (0,0,3)\) in the direction \((-1,0,-2)\text{.}\)
Solution.
First I need to make the direction a unit vector. The length of the direction is
\begin{equation*} |v| = \sqrt{(-1)^2 + (-2)^2} = \sqrt{5} \end{equation*}
so the unit vector is
\begin{equation*} u = \frac{v}{|v|} = \left( \frac{-1}{\sqrt{5}}, 0, \frac{-2}{\sqrt{5}} \right) \text{.} \end{equation*}
Then I calculate the gradient of \(f\text{,}\) taking the partials in each variable.
\begin{equation*} \nabla f = \left( yze^{xyz}, xze^{xyz} xye^{xyz} \right) \end{equation*}
Finally, the directional derivative is the dot product of gradient and the direction.
\begin{equation*} D_u f = \frac{1}{\sqrt{5}} (-yze^{xyz} - 2xye^{xyz}) \end{equation*}
Finally I eavluate this at the point.
\begin{equation*} D_u f (p) = \frac{0}{\sqrt{5}} = 0 \end{equation*}

Activity 4.6.5.

Calculate the directional derivative of \(f(x,y) = \frac{x^2 + y^2}{1 + z^2}\) at the point \(p = (2,0,-2)\) in the direction \((1,1,1)\text{.}\)
Solution.
First I need to make the direction a unit vector. The length of the direction is
\begin{equation*} |v| = \sqrt{1^2 + 1^2 +1^2} = 3 \end{equation*}
so the unit vector is
\begin{equation*} u = \frac{v}{|v|} = \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \text{.} \end{equation*}
Then I calculate the gradient of \(f\text{,}\) taking the partials in each variable.
\begin{equation*} \nabla f = \left( \frac{2x}{1 + z^2}, \frac{2y}{1 + z^2} \frac{-2z(x^2 +y^2)}{(1 + z^2)^2} \right) \end{equation*}
Finally, the directional derivative is the dot product of gradient and the direction.
\begin{equation*} D_u f = \frac{1}{\sqrt{3}} \left( \frac{2x}{1 + z^2} + \frac{2y}{1+ x^2} - \frac{2z(x^2 + y^2)}{(1+z^2)^2} \right) \end{equation*}
Finally I eavluate this at the point.
\begin{equation*} D_u f (p) = \frac{1}{\sqrt{3}} \left( \frac{4}{5} + 0 + \frac{16}{25} \right) = \frac{32}{25} \sqrt{3} \end{equation*}

Subsection 4.6.2 Chain Rule

Activity 4.6.6.

Calculate the derivative of the scalar field \(f(x,y) = 4x + 3y \) along the parametric curve \(\gamma(t) = (t^2 + t - 1, t^3 - 2t + t + 1) \) in two ways: first by using the chain rule and secondly by calculating the whole composition and doing a simple single-variable \(t\) derivative. Check that both answers are the same.
Solution.
To do the chain rule, first I need the partials of f.
\begin{equation*} \frac{\del f}{\del x} = 4 \hspace{1cm} \frac{\del f}{\del y} = 3 \end{equation*}
Then I need the tangent to the curve.
\begin{equation*} \gamma^\prime (t) = \left( 2t + 1, 3t^2 - 2t + 1 \right) \end{equation*}
Then I put these together into the chain rule.
\begin{equation*} \frac{d}{dt} f(\gamma(t)) = \frac{df}{dx} \frac{dx}{dt} + \frac{df}{dy} \frac{dy}{dt} = 4(2t + 1) + 3 (3t^2 - 2t + 1) = 9t^2 + 2t + 7 \end{equation*}
Then I substitute the \(x\) and \(y\) with the components of the curve, or I would do if they were present. Since the partials are constant here, there is no such substitution to do
More directly, I calculate the whole composition by substituting the components of the curve into the field.
\begin{equation*} f(\gamma(t)) = (t^2 + t - 1) + 3 (t^3 - 2t^2 + t + 1)= 3t^3 + t^2 + 7t - 1 \end{equation*}
Then I just calcualte the \(t\) derivative of this single-variable function.
\begin{equation*} \frac{d}{dt} f(\gamma(t)) = \frac{d}{df} 3t^3 + t^2 + 7t - 1 = 9t^2 + 2t + 7 \end{equation*}
Both results do produce the same derivative.

Activity 4.6.7.

Calculate the derivative of the scalar field \(f(x,y) = x^2 - y^2 \) along the parametric curve \(\gamma(t) = (\cos t, \sin t) \) in two ways: first by using the chain rule and secondly by calculating the whole composition and doing a simple single-variable \(t\) derivative. Check that both answers are the same.
Solution.
To do the chain rule, first I need the partials of f.
\begin{equation*} \frac{\del f}{\del x} = 2x \hspace{1cm} \frac{\del f}{\del y} = -2y \end{equation*}
Then I need the tangent to the curve.
\begin{equation*} \gamma^\prime (t) = \left( -\sin t, \cos t \right) \end{equation*}
Then I put these together into the chain rule.
\begin{equation*} \frac{d}{dt} f(\gamma(t)) = \frac{df}{dx} \frac{dx}{dt} + \frac{df}{dy} \frac{dy}{dt} = 2x (-\sin t) - 2y (\cos t) \end{equation*}
Then I substitute the \(x\) and \(y\) with the components of the curve.
\begin{equation*} = -2 (\cos t) \sin t) = 2 \sin t (\cos t) = -4 \sin t \cos t = -4 \sin (2t) \end{equation*}
More directly, I calculate the whole composition by substituting the components of the curve into the field.
\begin{equation*} f(\gamma(t)) = \cos^2 t - \sin^2 t \end{equation*}
Then I just calcualte the \(t\) derivative of this single-variable function.
\begin{align*} \frac{d}{dt} f(\gamma(t)) \amp = \frac{d}{dt} \cos^2 t - \sin^2 t \\ \amp = -2 \cos t \sin t - 2 \sin t \cos t = -4 \sin t \cos t = -2 \sin (2t) \end{align*}
Both results do produce the same derivative.

Activity 4.6.8.

Calculate the derivative of the scalar field \(f(x,y) = \frac{1}{x^2 + y^2 + 1} \) along the parametric curve \(\gamma(t) = (3t - 2, t + 4) \) in two ways: first by using the chain rule and secondly by calculating the whole composition and doing a simple single-variable \(t\) derivative. Check that both answers are the same.
Solution.
To do the chain rule, first I need the partials of f.
\begin{equation*} \frac{\del f}{\del x} = \frac{-2x}{(x^2 + y^2 + 1)^2} \hspace{1cm} \frac{\del f}{\del y} = \frac{-2y}{(x^2 + y^2 + 1)^2} \text{.} \end{equation*}
Then I need the tangent to the curve.
\begin{equation*} \gamma^\prime (t) = (3, 1) \end{equation*}
Then I put these together into the chain rule.
\begin{align*} \frac{d}{dt} f(\gamma(t)) \amp = \frac{df}{dx} \frac{dx}{dt} + \frac{df}{dy} \frac{dy}{dt} =\\ \amp = \frac{-2x}{(x^2 + y^2 + 1)^2} (3) + \frac{-2y}{(x^2 + y^2 + 1)^2}(0) \\ \amp = \frac{-6x - 2y}{(x^2 +y^2 + 1)^2} \end{align*}
Then I substitute the \(x\) and \(y\) with the components of the curve.
\begin{equation*} = \frac{-18 t - 12 - 2t - 8}{((3t-2)^2 + (t_4)^2 + 1)^2} = \frac{-20t + 4}{(10t^2 - 4t + 21)^2} \end{equation*}
More directly, I calculate the whole composition by substituting the components of the curve into the field.
\begin{equation*} f(\gamma(t)) = \frac{1}{((3t-2)^2 + (t+4)^2 + 1)} = \frac{1}{(10t^2 - 4t + 21)} \end{equation*}
Then I just calcualte the \(t\) derivative of this single-variable function.
\begin{equation*} \frac{d}{dt} f(\gamma(t)) = \frac{d}{dt} \frac{1}{10t^2 - 4t + 21} = \frac{-20t + 4}{(10t^2 - 4t + 21)} \end{equation*}
Both results do produce the same derivative.

Activity 4.6.9.

Calculate the derivative of the scalar field \(f(x,y) = e^{-(x^2 + y^2)} \) along the parametric curve \(\gamma(t) = (t \cos t, t \sin t) \) in two ways: first by using the chain rule and secondly by calculating the whole composition and doing a simple single-variable \(t\) derivative. Check that both answers are the same.
Solution.
To do the chain rule, first I need the partials of f.
\begin{equation*} \frac{\del f}{\del x} = -2xe^{-(x^2 + y^2)} \hspace{1cm} \frac{\del f}{\del y} = -2ye^{-(x^2 + y^2)} \text{.} \end{equation*}
Then I need the tangent to the curve.
\begin{equation*} \gamma^\prime (t) = \left( cos t - t \sin t, \sin t + t \cos t \right) \end{equation*}
Then I put these together into the chain rule.
\begin{align*} \frac{d}{dt} f(\gamma(t)) \amp = \frac{df}{dx} \frac{dx}{dt} + \frac{df}{dy} \frac{dy}{dt} .\\ \amp = 2xe^{-(x^2 + y^2)} (cos t - t \sin t) - 2ye^{-(x^2 + y^2)} (\sin t + t \cos t) \end{align*}
Then I substitute the \(x\) and \(y\) with the components of the curve.
\begin{align*} \amp = \left( e^{-(t^2 \cos^2 t + t^2 \sin^2 t)} \right) \left( -2(t \cos t)(cos t - t \sin t) - 2 (t \sin t) (\sin t + t \cos t) \right) \\ \amp = e^{-t^2} \left( -2t \cos^2 t + 2t^2 \sin t \cos t - 2t\sin^2 t - 2t^2 \sin t \cos t \right) \\ \amp = e^{-t^2} (-2t(\cos^2 t + \sin^2 t)) = -2te^{-t^2} \end{align*}
More directly, I calculate the whole composition by substituting the components of the curve into the field.
\begin{equation*} f(\gamma(t)) = e^{-(t^2 \cos^2 t + t^2 \sin^2 t)} = e^{-t^2} \end{equation*}
Then I just calcualte the \(t\) derivative of this single-variable function.
\begin{equation*} \frac{d}{dt} f(\gamma(t)) = \frac{d}{dt} e^{-t^2} = -2t e^{-t^2} \end{equation*}
Both results do produce the same derivative.

Activity 4.6.10.

Consider the gravitation potential defined from the videos.
\begin{equation*} P(x,y,z) = \frac{-GmM}{\sqrt{x^2 + y^2 + z^2}} \end{equation*}
Calcualte the change in potential energy along the curve \(\gamma(t) = 3 \cos t, 3 \sin t, 0) \text{.}\) Describe the curve and give reasons for the increase or decrease of potential energy along this curve (using the fact that potential energy decreases to \(-\infty\) near the origin and decreases away from the origin, to a limit of \(0\) at infinite distance).
Solution.
To do the chain rule, first I need the partials of f.
\begin{align*} \frac{\del f}{\del x} \amp = \frac{GmMx}{(x^2 + y^2 + z^2)^\frac{3}{2}} \\ \frac{\del f}{\del y} \amp = \frac{GmMy}{(x^2 + y^2 + z^2)^\frac{3}{2}} \\ \frac{\del f}{\del z} \amp = \frac{GmMz}{(x^2 + y^2 + z^2)^\frac{3}{2}} \text{.} \end{align*}
Then I need the tangent to the curve.
\begin{equation*} \gamma^\prime (t) = \left( -3 \sin t, 3 \cos t, 0 \right) \end{equation*}
Then I put these together into the chain rule.
\begin{align*} \frac{d}{dt} P(\gamma(t)) = \amp \frac{GmMx}{(x^2 + y^2 + z^2)^\frac{3}{2}} (-3 \sin t) + \frac{GmMy}{(x^2 + y^2 + z^2)^\frac{3}{2}} (3 \cos t) \\ \amp + \frac{GmMz}{(x^2 + y^2 + z^2)^\frac{3}{2}} (0) \end{align*}
Then I substitute the \(x\) and \(y\) with the components of the curve.
\begin{align*} \amp = \frac{GmM3 \cos t}{(9\cos^2 t + 9\sin^2 t + 0)^\frac{3}{2}} (-3 \sin t) + \frac{GmM3 \sin t}{(9\cos^2 t + 9\sin^2 t + 0)^\frac{3}{2}} (3 \cos t)\\ \amp + \frac{GmM0}{(9\cos^2 t + 9\sin^2 t + 0)^\frac{3}{2}} (0)\\ \amp = \frac{GmM9((\cos t (-\sin t) + (\sin t)(\cos t)}{27} = 0 \end{align*}
These is no change in potential energy along this path. This is a circule path in the \(xy\) plane. The potential is circulaly symmetric. Moving in a circle (an orbit, really) around the central object, never getting father away or close, should keep potential energy fixed, as indeed it does.

Activity 4.6.11.

Consider the gravitation potential defined from the videos.
\begin{equation*} P(x,y,z) = \frac{-GmM}{\sqrt{x^2 + y^2 + z^2}} \end{equation*}
Calcualte the change in potential energy along the curve \(\gamma(t) = \left( \frac{1}{t} \cos t, \frac{1}{t} \sin t\right) \text{.}\) Describe the curve and give reasons for the increase or decrease of potential energy along this curve (using the fact that potential energy decreases to \(-\infty\) near the origin and decreases away from the origin, to a limit of \(0\) at infinite distance).
Solution.
To do the chain rule, first I need the partials of f.
\begin{align*} \frac{\del f}{\del x} \amp = \frac{GmMx}{(x^2 + y^2 + z^2)^\frac{3}{2}} \\ \frac{\del f}{\del y} \amp = \frac{GmMy}{(x^2 + y^2 + z^2)^\frac{3}{2}} \\ \frac{\del f}{\del z} \amp = \frac{GmMz}{(x^2 + y^2 + z^2)^\frac{3}{2}} \text{.} \end{align*}
Then I need the tangent to the curve.
\begin{equation*} \gamma^\prime (t) = \left( \frac{-1}{t^2} \cos t - \frac{1}{t} \sin t, \frac{-1}{t^2} \sin t + \frac{1}{t} \cos t, 0 \right) \end{equation*}
Then I put these together into the chain rule.
\begin{align*} \frac{d}{dt} P(\gamma(t)) = \amp \frac{GmMx}{(x^2 + y^2 + z^2)^\frac{3}{2}} (-3 \sin t) + \frac{GmMy}{(x^2 + y^2 + z^2)^\frac{3}{2}} (3 \cos t) \\ \amp + \frac{GmMz}{(x^2 + y^2 + z^2)^\frac{3}{2}} (0) \end{align*}
Then I substitute the \(x\) and \(y\) with the components of the curve.
\begin{align*} \amp = \frac{GmM \frac{1}{t} \cos t}{ \left( \frac{ \cos^2 t}{t^2} + \frac{\sin ^2 t}{t^2} \right)^{\frac{3}{2}}} \left( \frac{-1}{t^2} \cos t - \frac{1}{t} \sin t \right) \\ \amp + \frac{GmM \frac{1}{t} \sin t}{\left( \frac{\cos^2 t}{t^2} + \frac{\sin^2 t}{t^2} \right)^{\frac{3}{2}}} \left( \frac{-1}{t^2} \sin t + \frac{1}{t} \cos t \right) \\ \amp = \frac{GmM \frac{1}{t} \cos t}{\left(\frac{1}{t^2} \right)^{\frac{3}{2}}} + \left( \frac{-1}{t^2} \cos t - \frac{1}{t} \sin t \right) \\ \amp + \frac{GmM \frac{1}{t} \sin t}{\left( \frac{1}{t^2} \right)^{\frac{3}{2}}} \left( \frac{-1}{t^2} \sin t + \frac{1}{t} \cos t \right) \\ \amp = GmM (\cos^2 t - t \sin t \cos t - \sin^2 t + t \sin t \cos t) = - GmM \end{align*}
The potential energy has a constant rate of decrease. This path is a spiral in the \(xy\) plane with a reciprocal delay in radius, spiralling into the centre. It is approaching the origin, so the potential energy should always increase. The fact that the potential energy increases linearly reflects that the potential energy is related to \(1/t\) but the radius is also decreasing as \(1/t\text{:}\) the alignment of these two means that the rate of increease of potential energy should be constant.

Activity 4.6.12.

Consider the gravitation potential defined from the videos.
\begin{equation*} P(x,y,z) = \frac{-GmM}{\sqrt{x^2 + y^2 + z^2}} \end{equation*}
Calcualte the change in potential energy along the curve \(\gamma(t) = (3 \cos t, 4 \sin t, 4 \cos t) \text{.}\) Describe the curve and give reasons for the increase or decrease of potential energy along this curve (using the fact that potential energy decreases to \(-\infty\) near the origin and decreases away from the origin, to a limit of \(0\) at infinite distance).
Solution.
To do the chain rule, first I need the partials of f.
\begin{align*} \frac{\del f}{\del x} \amp = \frac{GmMx}{(x^2 + y^2 + z^2)^\frac{3}{2}} \\ \frac{\del f}{\del y} \amp = \frac{GmMy}{(x^2 + y^2 + z^2)^\frac{3}{2}} \\ \frac{\del f}{\del z} \amp = \frac{GmMz}{(x^2 + y^2 + z^2)^\frac{3}{2}} \text{.} \end{align*}
Then I need the tangent to the curve.
\begin{equation*} \gamma^\prime (t) = \left( -3 \sin t, 5 \cos t, -4 \sin t \right) \end{equation*}
Then I put these together into the chain rule.
\begin{align*} \frac{d}{dt} P(\gamma(t)) = \amp \frac{GmMx}{(x^2 + y^2 + z^2)^\frac{3}{2}} (-3 \sin t) + \frac{GmMy}{(x^2 + y^2 + z^2)^\frac{3}{2}} (3 \cos t) \\ \amp + \frac{GmMz}{(x^2 + y^2 + z^2)^\frac{3}{2}} (0) \end{align*}
Then I substitute the \(x\) and \(y\) with the components of the curve.
\begin{align*} \amp = \frac{GmM}{(25 \cos^2 t + 25 \sin^2 t)^{\frac{3}{2}}} \left( 3 \cos t (-3 \sin t) + 5 \sin t (5 cos t) + (4 \cos t) (-4 \sin t) \right) \\ \amp = \frac{GmM}{(25)^{\frac{3}{2}}} (-9 \cos t \sin t + 2t \sin t \cos t - 16 \sin t \cos t) \\ \amp = \frac{GmM}{125} (\sin t \cos t)(-9 + 25 - 16) = 0 \end{align*}
The fact that there is, again, no change of potential energy here is surprising. That means that the distance to the origin must never change. This combination of sines and cosines must give a path that is restricted to a fixed sphere. The distance at, say, t=0 is 5 units, so the path is restricted to the sphere of radius 5. What exactly this path does on the sphere of radius 5: whether it makes a simple loop or traces a more complicated repeated figure over the sphere, is, I think, a bit difficult to say from the fnctions themselves. If you graph this by computer, it is, in fact, a great circle on the sphere, just going around in a circular orbit, but again, without computer assistance that’s a pretty difficult thing to see in the curve expressions themselves.

Subsection 4.6.3 Tangent Planes

Activity 4.6.13.

Calculate the equation of the tangent plane to the scalar field \(f(x,y) = 3y - 4x^2\) at the point \(p = (0,0) \text{.}\)
Solution.
The equation to the tangent line has this form.
\begin{equation*} z - f(a,b) = f_x(a,b) (x-a) + f_y(a,b) (y-b) \end{equation*}
I need to calculate the values of the function and its derivatives and put them into this form.
\begin{align*} f(0,0) \amp = 3(0) - 4(0)^2 = 0 \\ f_x(x,y) \amp = -8x \\ f_x(0,0) \amp = 0 \\ f_y(x,y) \amp = 3 \\ f_y(0,0) \amp = 3 \end{align*}
Then I put all the data into the equation.
\begin{equation*} z - 0 = 0 (x-0) + 3 (y-0) \implies z = 3y \end{equation*}
This is the equation of the tangent plane at the point.

Activity 4.6.14.

Calculate the equation of the tangent plane to the scalar field \(f(x,y) = e^{x^2 + y^2}\) at the point \(p = (0,0) \text{.}\)
Solution.
The equation to the tangent line has this form.
\begin{equation*} z - f(a,b) = f_x(a,b) (x-a) + f_y(a,b) (y-b) \end{equation*}
I need to calculate the values of the function and its derivatives and put them into this form.
\begin{align*} f(0,0) \amp = 1 \\ f_x(x,y) \amp = 2xe^{x^2 + y^2} \\ f_x(0,0) \amp = 0 \\ f_y(x,y) \amp = 2ye^{x^2 + y^2} \\ f_y(0,0) \amp = 0 \end{align*}
Then I put all the data into the equation.
\begin{equation*} z - 1 = 0 (x-0) + 0 (y-0) \implies z = 1 \end{equation*}
This is the equation of the tangent plane at the point.

Activity 4.6.15.

Calculate the equation of the tangent plane to the scalar field \(f(x,y) = \frac{x^2 + 1}{y^2 - 4}\) at the point \(p = (=3,3)\text{.}\)
Solution.
The equation to the tangent line has this form.
\begin{equation*} z - f(a,b) = f_x(a,b) (x-a) + f_y(a,b) (y-b) \end{equation*}
I need to calculate the values of the function and its derivatives and put them into this form.
\begin{align*} f(-3,3) \amp = \frac{9}{5} \\ f_x(x,y) \amp = \frac{2x}{y^2 - 4} \\ f_x(-3,3) \amp = \frac{-6}{5} \\ f_y(x,y) \amp = \frac{-(2y)(x^2 + 1)}{(y^2 - 4)^2} \\ f_y(-3,3) \amp = \frac{-12}{5} \end{align*}
Then I put all the data into the equation.
\begin{equation*} z - \frac{9}{5} = \frac{-6}{5} (x+3) + \frac{-12}{5} (y-3) \end{equation*}
This simplifies to
\begin{equation*} 27 = 6x + 12y + 5z \end{equation*}
This is the equation of the tangent plane at the point.

Subsection 4.6.4 Conceptual Review Questions

  • What is a direction derivative and how does it vary from a partial and a gradient?
  • How does a tangent (hyper) plane extend the idea of a tangent line?
  • Why does the composition of a scalar field with a parametric curve make sense?
  • Why is the generalization of the derivative so complicated? Why is the linear approximation a good generalization?
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