Multivariable Calculus – Change of Bound Help!

I am having trouble following these steps in a reading on multivariable calculus.

Due to a change of variables: 

$ \displaystyle\int_t^T \int_t^s \theta_v dv \, ds = \int_t^T \int_s^T \theta_s dv \, ds$

Could anyone explain how you get the right equation? What is the ‘change of variable’ that is applied? Thanks!! Also $ t \leq T$ is given.

Calculus 1 Courses’s Optimization Homework Problem, Produces Strangely Complex Value.

I am a Calculus 1 student and I have an optimization word-problem that is giving me a lot of trouble.

It has two variables. I have found the value for $ y$ , but when I plugged it into the equation and tried to solve for the $ x$ I couldn’t find it’s value. I used Symbolab to solve it, but it came up with a decimal number that’s extremely complicated when written as a fraction. My professor has given us very complicated problems before, but the complexity of this number is such that I feel like it’s very likely I did something wrong.

I have checked other parts of my work with Symbolab and I am still not sure where I went wrong, but I would really appreciate it if you would take a look and determine if there are any parts that don’t look right to you.

An oil refinery is located on the north bank of a straight river which is $ 2$ km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river $ 6$ km east of the refinery. The cost of laying pipe is $ $ 400,000$ per km over land to a point $ P$ on the north bank and $ $ 800,000$ per km under the river to the tanks. To minimize the cost of the pipeline, where should $ P$ be located?

$ P=$ The area where the pipeline enters the river.

$ x=$ The horizontal distance between the oil refinery and the storage tanks.

$ y=$ The euclidean distance between $ P$ and the storage tanks.

The Pythagorean theorem states $ 2^2+(6-x)^2=y^2$ .

The cost of the pipeline is $ C = 400,000x+800,000y$ .

finding y:

$ $ 4+(6-x)^2 = y^2 \to y= \pm \sqrt{4+(6-x)^2}$ $

Differentiating $ C = 400,000x+800,000y$ :

$ $ \frac{d}{dx}\sqrt{4+(6-x)^2}=\frac{1}{2\sqrt{4+(6-x)^2}}\cdot-2(6-x)=\frac{-(6-x)}{\sqrt{4+(6-x)^2}}$ $

$ $ \frac{d}{dx}800,000\sqrt{4+(6-x)^2}=[\frac{-(6-x)}{\sqrt{4+(6-x)^2}}\cdot800,000] = \frac{-800,000(6-x)}{\sqrt{4+(6-x)^2}}$ $

$ $ \frac{d}{dx}400,000x+800,000\sqrt{4+(6-x)^2}=400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}}$ $

Setting $ 400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}} = 0$ and solving for $ x$ .

$ $ x=4.84530..$ $

I’m not completely sure how to write the fraction out with math notation here because a single square root seems to cover part of the numerator and all of the denominator, but you can see it if you plug $ 400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}} = 0$ into Symbolab’s “solve for” calculator.

Vectors (calculus 3)

I was reviewing Calculus 3 for my upcoming analysis class and was looking at Vector Functions. I just have a question about this specific one for when I look at $ t = 2$ .

$ \vec{r}(t) = \sqrt{t+2}\hat{i} + 3t\hat{j} + \frac{1}{t+1}\hat{k}$

My question is, does $ \vec{r}(2) = 2\hat{i} + 6\hat{j} + \frac{1}{3}\hat{k}$ only, or is it ALSO $ \vec{r}(2) = -2\hat{i} + 6\hat{j} + \frac{1}{3}\hat{k}$ .

I guess the general question is, can a Vector Function give 2 different vectors for one value of the parameter?

Is there such thing as directional derivative in single variable calculus?

In multivariable calculus, the directional derivative is described in the following way:

While hiking in rugged terrain, you might think of your altitude at the point given by longitude x and latitude y as defining a function $ f (x, y)$ . Although you won’t have a handy formula for this function, you can say more about this function than you might expect. If you face due east (in the direction of the positive x-axis), the slope of the terrain is given by the partial derivative $ {∂ f\over ∂x} (x, y)$ . Similarly, facing due north, the slope of the terrain is given by $ {∂ f\over ∂y} (x, y)$ . However, how would you compute the slope in some other direction, say north-by-northwest? In this section, we develop the notion of directional derivative, which will answer this question. (Calculus, Early Transcendental Function, 4th ed. Smith and Minton)

Usually the diagram that is given is something like this:enter image description here

all of which seem to make it very clear that this is a concept in multivariable calculus, such as in $ R^3$ .

I am wondering if such a concept exists in single variable calculus as well.

If I have a function such as $ x^2$ , we know how to find the rate of change at a point. But what about the rate of change in the direction of a certain $ R^2$ vector? Is there a reason why such a concept would not make sense? And if it does, what would be the formula for it? Would it be simply some modification of the multivariable formula?

Expressiveness of lambda calculus without null variable application

If the lambda calculus is restricted to disallow applications of null variables (for example, something like $ \lambda x.E[x:=\emptyset]\rightarrow\lambda x.E$ ), what is the effect on the class of expressions that can be generated? Given that sub-expressions may still contract (just not disappear entirely) is it more expressive than a non-contracting/context-sensitive grammar?

I’m guessing this is a basic question that I haven’t found the right search terms for yet. Thanks for any guidance.

Why ‘let’ can not be reduce to a lambda application in (extended) Calculus of Constructions

I do not understand the difference highlighted in the chapter 2.5 of the book Theorem Proving in Lean:

Notice that the meaning of the expression let a := t1 in t2 is very similar to the meaning of (λ a, t2) t1, but the two are not the same. In the first expression, you should think of every instance of a in t2 as a syntactic abbreviation for t1. In the second expression, a is a variable, and the expression λ a, t2 has to make sense independently of the value of a. The let construct is a stronger means of abbreviation, and there are expressions of the form let a := t1 in t2 that cannot be expressed as (λ a, t2) t1. As an exercise, try to understand why the definition of foo below type checks, but the definition of bar does not.

def foo := let a := nat  in λ x : a, x + 2 def bar := (λ a, λ x : a, x + 2) nat 

What is the explanation behind this?

Matrix Calculus: Scalar by Matrix Derivative

I’ve been self studying Matrix Calculus for awhile, I had no problem understanding the gradient, the Jacobian and the Hessian and went through multiple exercises and examples for each, I got stuck when trying to understand and differentiate a scalar $ y$ function of a $ p×q$ matrix $ \textbf{X}$ .
if the scalar $ f$ function of a vector $ \textbf{x}$ takes the following form for example: $ $ x_1 – 3x_2 + 4x_3^3$ $ Then how would the scalar $ y$ function of a vector $ \textbf{X}$ look like?

Calculus: Finding an unknown coefficient in continuous equations

I am watching a video trying to understand continuity in calculus. The author first starts with the $ ax – 4, x < 1$ , and we are trying to find the value of “a”.

The author subsequently derives $ ax – 4$ as $ 7x – 4$ , where $ a = 7$ .

Why did he come to a conclusion where $ a$ is $ 7$ , where $ x < 1$ ?

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