Mathematics, 19.11.2020 14:00 kingoc321

A curve is such that (dy)/(dx) = 12/(2x + 1) ^ 2 and P(1, 5) is a point on the curve . (i) The normal to the curve at P crosses the x-axis at Q. Find the coordinates of Q (ii) Find the equation of the curve. (iii) A point is moving along the curve in such a way that the x-coordinate is increasing at a constant rate of 0.3 units per second. Find the rate of increase of the y-coordinate when x=1

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Answer from: jumana3

(i) The coordinates of the point Q is (x, y) = (21, 0) .

(ii) The equation of the curve is $y = 6\cdot \ln (2\cdot x + 1)-1.592$ .

(iii) The rate of increase of the y-coordinate is 1.2 units per second.

(i) By analytical geometry, we know that the slope of a line normal of a curve ( $m_{\perp}$ ) by the following expression:

$m_{\perp} = - \frac{1}{\frac{dy}{dx} }$ (1)

Where $\frac{dy}{dx}$ is the slope of the line tangent to the curve.

If we know that x = 1 , then the slope of the normal to the curve is:

$\frac{dy}{dx} = \frac{12}{2\cdot (1)+1}$

$\frac{dy}{dx} = 4$

And the slope of a line normal of a curve is:

$m_{\perp} = -\frac{1}{4}$

By definition of secant curve we calculate the remaining coordinate of the point Q:

$m_{\perp} = \frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}$ (2)

Where:

$x_{P}$ , $x_{Q}$ - x-coordinates of points P and Q. $y_{P}, y_{Q}$ - y-coordinates of points P and Q.

If we know that $(x_{P}, y_{P}) = (1, 5)$ , $(x_{Q}, y_{Q}) = (x, 0)$ and $m_{\perp} = -\frac{1}{4}$ , then the remaining coordinate of point Q is:

$-\frac{1}{4} = \frac{0-5}{x-1}$

$-\frac{1}{4}\cdot (x-1) = -5$

$-\frac{1}{4}\cdot x +\frac{1}{4} = -5$

$\frac{1}{4}\cdot x = \frac{21}{4}$

x = 21

The coordinates of the point Q is (x, y) = (21, 0) .

(ii) The equation of the curve is derived by separating the variables and integrating the resulting expression:

$dy = \frac{12}{2\cdot x + 1}\,dx$

$\int\, dy = 6\int {\frac{2\,dx}{2\cdot x + 1} }$

$\int \, dy = 6\int {\frac{du}{u} }$

$y = 6\cdot \ln u$

$y = 6\cdot \ln (2\cdot x + 1)+C$

And the integration constant is found for (x,y) = (1, 5) :

$5 = 6\cdot \ln(2\cdot 1 + 1) + C$

C = -1.592

The equation of the curve is $y = 6\cdot \ln (2\cdot x + 1)-1.592$ .

(iii) The rate of increase of the y-coordinate by using the differential equation presented in statement and the definition of rate of change:

$\frac{dy}{dt} = \left(\frac{12}{2\cdot x +1}\right)\cdot \frac{dx}{dt}$ (3)

If we know that $\frac{dx}{dt} = 0.3$ and x = 1 , then the rate of increase of the y-coordinate is:

$\frac{dy}{dt} = \left[\frac{12}{2\cdot (1)+1} \right]\cdot (0.3)$

$\frac{dy}{dt} = 1.2$

The rate of increase of the y-coordinate is 1.2 units per second.

To learn more on differential equations, we kindly invite to check this verified question: link

Answer from: sonnyboi2305

i). The coordinates of Q is equal to:

(x,y) = (21, 0)

ii). The equation of the curve is:

y = 6. In(2.x + 1) - 1.592

iii). The rate of increase for y-coordinate when x = 1:

1.2 units per second

i). Using analytical geometry,

and are the P and Q's x-coordinates.

and are the P and Q's y-coordinates.

Since (, ) = (1,5) and ( , ) = (x, 0) and m⊥ = - 1/4, the left coordinate of Q's point:-

-1/4 = (0-5)/(x - 1)

∵ x = 21

Thus, the coordinates of y = (21, 0)

ii). Through isolating the variables and employing integration:-

$dy = \frac{12}{(2.x + 1)} dx$

by integration and integration constant (1,5), the equation is:

y = 6. In(2.x + 1) - 1.592

iii). The rate of increase for y-coordinate:

dy/dt = $\frac{(12)}{(2.1 + 1)} .0.3$

∵ dy/dt = 1.2

Thus, the rate of increase for y-coordinate when x = 1 is 1.2 units per second.

Learn more about "Co-ordinates" here:

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Answer from: Quest

The answer is c because you turn 17.5% into a decimal which gives you .175 then you times that by $361.35 which gives you 63.23625 but then you round that which gives you 63.24. then you at 63.24 pulse 361.35 and you round it which gives you $425

Answer from: Quest

1. minimum

2. 6400

3. eight

hope this !

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A curve is such that (dy)/(dx) = 12/(2x + 1) ^ 2 and P(1, 5) is a point on the curve . (i) The norma...

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