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Chapter 1: Vector Analysis Problem 1.1: Using the definitions in Eqs. 1.1 and 1.4, and appropriate diagrams, show that the dot product and cross product are distributive, a) when the three vectors are co-planar. b) in the general case.  Solution   Problem 1.2 Is the cross product associative? $$(\vec{A}\times \vec{B}) \times \vec{C} \overset{?}{=} \vec{A}\times (\vec{B} \times \vec{C})$$ If so, prove it; if not, provide a counterexample (the simpler the better). Solution   Problem 1.3 Find the angle between the body diagonals of a cube.  Solution Problem 1.4 Use the cross product to find the components of the unit vector $\hat{n}$ perpendicular to the shaded plane in Fig. 1.11. Solution   Problem 1.5 Prove the BAC-CAB rule by writing out both sides in component form. Solution Problem 1.6 Prove that $$[\vec{A}\times (\vec{B}\times \vec{C})]+[\vec{B}\times (\vec{C}\times \vec{A})]+[\vec{C}\times (\vec{A}\times \vec{B})] = 0$$ Under what conditions does $\vec{A}...
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Chapter 1 Vector Analysis: Problem 1.12

Problem 1.12 The height of a certain hill (in feet) is given by
$$h(x, y) = 10(2xy − 3x^2 − 4y^2 − 18x + 28y + 12),$$
where $y$ is the distance (in miles) north, $x$ the distance east of South Hadley.
(a) Where is the top of the hill located?
(b) How high is the hill?
(c) How steep is the slope (in feet per mile) at a point 1 mile north and one mile
east of South Hadley? In what direction is the slope steepest, at that point?

Solution:

Before proceeding to the answer, I strongly encourage you to watch this amazing video to get better insight of gradient: https://youtu.be/QQPz3eXXgQI OR if you are something who wants to learn by reading here's a great article https://betterexplained.com/articles/vector-calculus-understanding-the-gradient/ .

I hope you must have watched or/and read the article before proceeding.

$$h(x,y) = 10(2xy-3x^2-4y^2-18x+28y+12)$$


First let's compute gradient of the function:

$$\vec{\nabla} h = \frac{\partial h}{\partial x}\hat{i}+\frac{\partial h}{\partial y}\hat{j}$$
$$\vec{\nabla} h = 10(2y-6x-18)\hat{i}+10(2x-8y+28)\hat{j}$$

(a) Finding where the top of the hill is:

Since, gradient points in the direction of maximum change and we are already at the maximum of the hill. Therefore, gradient is zero at the top.

$$10(2y-6x-18) = 0$$

$$10(2x-8y+28) = 0$$

Solving the above equations,

$$x=-2,y=3$$ 

Therefore, the top of the hill is located 3 miles north and 2miles west of South Hadley.

(b) To find the height of the hill simply solve the function at $(x, y)=(-2, 3)$.

$$h(-2, 3) = 10[2\times (-2)\times 3-3\times (-2)^2-4\times (3)^2-18\times (-2)+28\times 3+12] = 720$$

Therefore, the hill is 720 feet high.

(c) Steepness at a point 1 mile north and 1 mile east of South Hadley. That is magnitude of  gradient at $y=1, x=1$

$$\vec{\nabla} h(x, y) = 10(2y-6x-18)\hat{i}+10(2x-8y+28)\hat{j}$$
$$\vec{\nabla} h(1,1) = 10(2(1)-6(1)-18)\hat{i}+10(2(1)-8(1)+28)\hat{j}$$
$$\vec{\nabla} h(1,1) = -220\hat{i}+220\hat{j}$$  
$$|\vec{\nabla} h(1,1)| = 220\sqrt{2}$$  

Therefore, the steepness at the given point is $220\sqrt{2}\text{ feet/mile}$.

If you move in the direction of the gradient you will get the steepest slope. Hence, if you move at $45^{\circ}$ north west, you will get steepest slope at that point.


If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.

 

Comments

  1. Anonymous27 January 2024 at 03:39

    hi, thank you for posting the solved problems with explained steps! it really is helpful, to understand beyond just copying :)

    ReplyDelete
    Replies
    1. Sumit Gupta27 January 2024 at 06:29

      Thanks. You know comments like this makes my day.

      Delete

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