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:

link

Answer from: Quest

$given : \frac{3}{9} = \frac{x}{5}$

cross multiplying, we get :

⇒ (3) × (5) = (9) × (x)

⇒ 15 = 9x

Answer from: Quest

the height of the surface increases from the center out to the sides of the road.

step-by-step explanation:

as we move out we move away from the x-axis.

and as mentioned in the problem statement that height of a road increases when we move away from the center x.

hence proved that 3rd option

the height of the surface increases from the center out to the sides of the road.

is the correct option.

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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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