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

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

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

(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 ([tex]m_{\perp}[/tex]) by the following expression:

[tex]m_{\perp} = - \frac{1}{\frac{dy}{dx} }[/tex] (1)

Where [tex]\frac{dy}{dx}[/tex] is the slope of the line tangent to the curve.

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

[tex]\frac{dy}{dx} = \frac{12}{2\cdot (1)+1}[/tex]

[tex]\frac{dy}{dx} = 4[/tex]

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

[tex]m_{\perp} = -\frac{1}{4}[/tex]

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

[tex]m_{\perp} = \frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}[/tex] (2)

Where:

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

[tex]-\frac{1}{4} = \frac{0-5}{x-1}[/tex]

[tex]-\frac{1}{4}\cdot (x-1) = -5[/tex]

[tex]-\frac{1}{4}\cdot x +\frac{1}{4} = -5[/tex]

[tex]\frac{1}{4}\cdot x = \frac{21}{4}[/tex]

[tex]x = 21[/tex]

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

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

[tex]dy = \frac{12}{2\cdot x + 1}\,dx[/tex]

[tex]\int\, dy = 6\int {\frac{2\,dx}{2\cdot x + 1} }[/tex]

[tex]\int \, dy = 6\int {\frac{du}{u} }[/tex]

[tex]y = 6\cdot \ln u[/tex]

[tex]y = 6\cdot \ln (2\cdot x + 1)+C[/tex]

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

[tex]5 = 6\cdot \ln(2\cdot 1 + 1) + C[/tex]

[tex]C = -1.592[/tex]

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

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

[tex]\frac{dy}{dt} = \left(\frac{12}{2\cdot x +1}\right)\cdot \frac{dx}{dt}[/tex] (3)

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

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

[tex]\frac{dy}{dt} = 1.2[/tex]

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: https://brainly.com/question/25731911

i). The coordinates of Q is equal to:

[tex](x,y) = (21, 0)[/tex]

ii). The equation of the curve is:

[tex]y = 6. In(2.x + 1) - 1.592[/tex]

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

[tex]1.2 units[/tex] per second

i). Using analytical geometry,

[tex]xp[/tex] and [tex]xq[/tex] are the P and Q's x-coordinates.

[tex]yp[/tex] and [tex]yq[/tex] are the P and Q's y-coordinates.

Since ([tex]xp[/tex], [tex]yp[/tex]) = (1,5) and ( [tex]xq[/tex], [tex]yq[/tex]) = [tex](x, 0)[/tex] and m⊥ = - 1/4, the left coordinate of Q's point:-

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

∵[tex]x = 21[/tex]

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

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

[tex]dy = \frac{12}{(2.x + 1)} dx[/tex]

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

[tex]y = 6. In(2.x + 1) - 1.592[/tex]

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

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

∵ [tex]dy/dt[/tex] [tex]= 1.2[/tex]

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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D) It needs to be included as it forces the reader to consider only two dimensions, like the xy-plane. The two parallel lines would determine the plane and the transversal would intersect both lines. This would allow the proof to use the Angle Bisector Theorem.

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