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2 years ago

From first principles, find the indicated derivatives

Mathematics

1 answer:

LenaWriter [7]2 years ago

6 0

By definition of the derivative,

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{(s + h)^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{s^3+3s^2h+3sh^2+h^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\frac{3s^2h+3sh^2+h^3}2}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 \frac{3s^2h+3sh^2+h^3}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 (3s^2+3sh+h^2)$

$\displaystyle\frac{dr}{ds} = \frac{3s^2}2$

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seropon [69]

Answer:

The equation of the line in point-slope form is $y-21 = - \frac{3}{2}\cdot (x-14)$ .

Step-by-step explanation:

According to the statement, let $A(x,y) = (14,21)$ and $B(x,y) = (1.25\cdot x_{A},0.75\cdot y_{A})$ . The equation of the line in point-slope form is defined by the following formula:

$y-y_{A} = m\cdot (x-x_{A})$ (1)

Where:

$x_{A}$ , $y_{A}$ - Coordinates of the point A, dimensionless.

$m$ - Slope, dimensionless.

$x$ - Independent variable, dimensionless.

$y$ - Dependent variable, dimensionless.

In addition, the slope of the line is defined by:

$m = \frac{y_{B}-y_{A}}{x_{B}-x_{A}}$ (2)

If we know that $x_{A} = 14$ and $y_{A} = 21$ , then the equation of the line in point-slope form is:

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$x_{B} = 17.5$

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$y_{B} = 15.75$

From (2):

$m = \frac{15.75-21}{17.5-14}$

$m = -\frac{3}{2}$

By (1):

$y-21 = - \frac{3}{2}\cdot (x-14)$

The equation of the line in point-slope form is $y-21 = - \frac{3}{2}\cdot (x-14)$ .

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Answer:

$\large \boxed{\textbf{30.54 yr}}$

Step-by-step explanation:

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From first principles, find the indicated derivatives​

From first principles, find the indicated derivatives