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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2. If a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other line.





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A) It does not need to be included. By leaving it out of the statement, the reader considers three dimensions not just the xy-plane. The two parallel lines would determine a plane. However, the transversal could intersect only one of the lines. This would allow the proof to use the Corresponding Angles Theorem.


B) It does not need to be included. By leaving it out of the statement, the reader considers three dimensions not just the xy-plane. The two parallel lines would determine a plane. However, the transversal could intersect only one of the lines. This would allow the proof to use the Angle Bisector Theorem.


C) 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 Corresponding Angles Theorem.


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.