Chapter 9 · Differential Equations

Exercise 9.3 · Q18: Tangent slope twice that of the segment to $(-4, -3)$ ⁠, through $(-2, 1)$

Exercise 9.3Variables separable, curve from a slope condition · Sections 9.3 and 9.4.1

The question

18. At any point $(x, y)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point $(-4, -3)$ ⁠. Find the equation of the curve given that it passes through $(-2, 1)$ ⁠.

Exam answer

Let $(x, y)$ be any point on the curve. The slope of the segment joining it to $(-4, -3)$ is $\frac{y + 3}{x + 4}$ ( $x \ne -4$ ⁠). Given

\frac{dy}{dx} = \frac{2(y + 3)}{x + 4}

Separating the variables ( $y \ne -3$ ⁠),

\frac{dy}{y + 3} = \frac{2\,dx}{x + 4}

Integrating both sides,

\begin{aligned} \log|y + 3| &= 2\log|x + 4| + \log|C| \\ y + 3 &= C(x + 4)^2\quad\ldots(\text{i}) \end{aligned}

The curve passes through $(-2, 1)$ ⁠. Putting $x = -2$ ⁠, $y = 1$ in (i): $4 = 4C$ ⁠, so $C = 1$ ⁠.

$\therefore$ The equation of the curve is $(x + 4)^2 = y + 3$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Both slopes in the sentence can be written in symbols: the tangent's slope is $\frac{dy}{dx}$ ⁠, and the segment's slope comes from the two-point formula. "Twice" turns the sentence into a separable differential equation whose solutions are the parabolas $y + 3 = C(x + 4)^2$ with vertex $(-4, -3)$ ⁠. The point $(-2, 1)$ picks $C = 1$ ⁠.

Given, to find, and the plan

Given: At every point $(x, y)$ of the curve, slope of the tangent $= 2 \times$ slope of the segment to $(-4, -3)$ ⁠; the curve passes through $(-2, 1)$ ⁠.
To find: The equation of the curve.
Plan: (1) Slope of the segment. (2) Write the differential equation. (3) Separate. (4) Integrate, with the constant as $\log|C|$ ⁠. (5) Use the point. (6) Check, and draw the tangent and the segment at the given point.

Basics you need

1Slope through two points

The line through $(x_1, y_1)$ and $(x_2, y_2)$ ⁠, with $x_1 \ne x_2$ ⁠, has slope $\frac{y_2 - y_1}{x_2 - x_1}$ ⁠. The slope of the tangent at $(x, y)$ is $\frac{dy}{dx}$ ⁠. Basics: an equation draws an arrow at every point

2Variables separable

If $\frac{dy}{dx} = g(x)\,h(y)$ with $h(y) \ne 0$ ⁠, then $\int\frac{dy}{h(y)} = \int g(x)\,dx$ (Section 9.4.1). Lesson: separate, then integrate

3Log integral, and the constant as a log

$\int\frac{dx}{x + a} = \log|x + a| + C$ (Section 7.2, formula (xii)). Writing the constant as $\log|C|$ lets the logs combine: $2\log|u| + \log|C| = \log|Cu^2|$ ⁠. Basics: every method ends with an integral

4A curve through a point

The given point fixes the constant (Section 9.3). Basics: one constant, one condition

Solution

Slope of the segment

The segment joins the point of contact $(x, y)$ to $(-4, -3)$ ⁠. Its slope is
$\frac{y - (-3)}{x - (-4)} = \frac{y + 3}{x + 4}\qquad (x \ne -4).$
Two-point slope formula. At $x = -4$ the segment would be vertical (or a single point), with no slope, so the condition only makes sense for $x \ne -4$ ⁠.

Slope of a line through two points (Class XI). Basics: an arrow at every point
Write the differential equation

The slope of the tangent is $\frac{dy}{dx}$ ⁠, and it is twice the slope of the segment:
$\frac{dy}{dx} = \frac{2(y + 3)}{x + 4}.$
"Twice" means multiply by 2. The equation has order 1 and degree 1, and has the separable form $g(x)\,h(y)$ with $g(x) = \frac{2}{x + 4}$ and $h(y) = y + 3$ ⁠.

Slope of the tangent is $\frac{dy}{dx}$ (Chapter 6). Basics: y′ is the slope
Separate the variables

For $y \ne -3$ ⁠, divide by $y + 3$ and multiply by $dx$ ⁠:
$\frac{dy}{y + 3} = \frac{2\,dx}{x + 4}.$
Dividing by $h(y) = y + 3$ needs $y \ne -3$ ⁠. The line $y = -3$ also satisfies the equation (both sides 0), but it does not pass through $(-2, 1)$ ⁠, so nothing we need is lost.

Section 9.4.1: separate, then integrate. Lesson: separate, then integrate
Integrate both sides, with one constant
$\begin{aligned} \log|y + 3| &= 2\log|x + 4| + \log|C| \\ &= \log|C(x + 4)^2|, \end{aligned}$
so $|y + 3| = |C(x + 4)^2|$ ⁠, that is,
$y + 3 = C(x + 4)^2.$
Both integrals are logs. The two constants combine into one, written $\log|C|$ with $C \ne 0$ (it can still be any real number). Then $2\log|u| = \log u^2$ and $\log m + \log n = \log mn$ give one log on each side, and log is one to one. The sign $\pm$ is absorbed into $C$ ⁠. This is the general solution: parabolas with vertex $(-4, -3)$ ⁠.

Section 7.2, formula (xii); laws of logarithms. Basics: one constant, one condition
Use the point to find $C$

Put $x = -2$ ⁠, $y = 1$ ⁠: $1 + 3 = C(-2 + 4)^2$ ⁠, that is $4 = 4C$ ⁠, so $C = 1$ and
$(x + 4)^2 = y + 3.$
One condition fixes the one constant. The curve is the parabola $y = (x + 4)^2 - 3$ ⁠, whose vertex is the given point $(-4, -3)$ itself.

Section 9.3: a particular solution gives the constant a value. Lesson: general and particular solutions
Check, and see the two slopes at $(-2, 1)$

From $y = (x + 4)^2 - 3$ ⁠: $\frac{dy}{dx} = 2(x + 4)$ ⁠, and $\frac{2(y + 3)}{x + 4} = \frac{2(x + 4)^2}{x + 4} = 2(x + 4)$ for $x \ne -4$ ⁠. At $(-2, 1)$ ⁠: $(-2 + 4)^2 = 4 = 1 + 3$ ⁠.

The parabola $(x + 4)^2 = y + 3$ ⁠, drawn with equal scales. At $P(-2, 1)$ the tangent (rose) has slope 4 and the segment to $Q(-4, -3)$ (dashed) has slope $\frac{1 + 3}{-2 + 4} = 2$ ⁠: the tangent is twice as steep. The hollow dot marks the vertex $Q$ ⁠, where the segment has no slope.

Both sides of the equation agree at every point with $x \ne -4$ ⁠, and the point lies on the curve. The condition holds on the whole parabola except its vertex $Q(-4, -3)$ ⁠; the piece through $(-2, 1)$ ⁠, on an interval containing $x = -2$ ⁠, is the right half, $x > -4$ ⁠.

A solution makes L.H.S. = R.H.S. Lesson: general and particular solutions

Result

The curve is the parabola $(x + 4)^2 = y + 3$ (the condition holds at every point except the vertex $(-4, -3)$ ⁠; the part through $(-2, 1)$ has $x > -4$ ⁠). (Textbook answer: $(x + 4)^2 = y + 3$ ⁠.)

Check it with numbers

At $x = -1$ ⁠: $y = 3^2 - 3 = 6$ ⁠. Tangent slope $2(x + 4) = 6$ ⁠; segment slope $\frac{6 + 3}{-1 + 4} = 3$ ⁠; and $6 = 2 \times 3$ ⁠.

At $P(-2, 1)$ from the curve itself: $y(-1.999) = 1.004001$ and $y(-2.001) = 0.996001$ ⁠, so

\frac{1.004001 - 0.996001}{0.002} = 4,

twice the segment slope 2.

Common mistakes

Sign slips in the segment slope: $y - (-3) = y + 3$ and $x - (-4) = x + 4$ ⁠.
Putting the 2 on the wrong side: the tangent's slope is twice the segment's, so $\frac{dy}{dx} = 2 \times \frac{y + 3}{x + 4}$ ⁠, not $\frac{1}{2}\times$ ⁠.
Writing $y + 3 = (x + 4)^2 + C$ ⁠. A constant added to logs becomes a factor: $y + 3 = C(x + 4)^2$ ⁠.

What to take away

Turn each slope in a sentence into symbols: tangent $\frac{dy}{dx}$ ⁠, segment by the two-point formula.
$\frac{dy}{y - b} = k\,\frac{dx}{x - a}$ gives $y - b = C(x - a)^k$ when $k$ is a whole number (here $k = 2$ ⁠).
Check the answer against the words: here the tangent at $(-2, 1)$ is twice as steep as the segment.

Every fact used, and where it comes from

Fact	Used in	Source
Slope through two points; slope of the tangent is $\frac{dy}{dx}$	Steps 1, 2	Class XI; Chapter 6 (Basics)
Variables separable: separate, then integrate	Step 3	Section 9.4.1 (Lesson)
$\int\frac{dx}{x + a} = \log\|x + a\| + C$ ⁠; laws of logarithms	Step 4	Section 7.2, formula (xii) (Basics)
A particular solution gives the constant a value	Step 5	Section 9.3 (Lesson)
A solution makes L.H.S. = R.H.S.	Step 6	Section 9.3 (Lesson)

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Exercise 9.3 · Q18: Tangent slope twice that of the segment to (−4,−3)(-4, -3)(−4,−3)⁠, through (−2,1)(-2, 1)(−2,1)

Basics you need

Solution

Slope of the segment

Write the differential equation

Separate the variables

Integrate both sides, with one constant

Use the point to find CCC

Check, and see the two slopes at (−2,1)(-2, 1)(−2,1)

Every fact used, and where it comes from

Exercise 9.3 · Q18: Tangent slope twice that of the segment to $(-4, -3)$ ⁠, through $(-2, 1)$

Use the point to find $C$

Check, and see the two slopes at $(-2, 1)$