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2 years ago

Consider the graph of the quadratic function. Which

1 answer:

6 0

Using the rate of change concept, it is found that it will be negative on the interval in which the function is decreasing.

<h3>What is the rate of change of a function?</h3>

The rate of change, considering two points, is given by the <u>change in the output divided by the change in the input</u>.

Hence:

If the rate is negative, the function is decreasing.

If the rate is positive, the function is increasing.

More can be learned about the rate of change of a function at brainly.com/question/24313700

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2 years ago

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2 years ago

Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) t

pashok25 [27]

Direct computation:

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$\vec r(t)=(x(t),y(t))=(\cos t,\sin t)$

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$\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)$

with $0\le t\le1$ . Then

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)$

$=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt$

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Using the fundamental theorem of calculus:

The integral can be written as

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}$

If there happens to be a scalar function $f$ such that $\vec F=\nabla f$ , then $\vec F$ is conservative and the integral is path-independent, so we only need to worry about the value of $f$ at the path's endpoints.