Elements of the Differential and Integral Calculus/Chapter XI

CHAPTER XI

CHANGE OF VARIABLE

95. Interchange of dependent and independent variables. It is sometimes desirable to transform an expression involving derivatives of y with respect to x into an equivalent expression involving instead derivatives of x with respect to y. Our examples will show that in many cases such a change transforms the given expression into a much simpler one. Or perhaps x is given as an explicit function of y in a problem, and it is found more convenient to use a formula involving $\tfrac{dx}{dy}$, $\tfrac{d^2 x}{dy^2}$, etc., than one involving $\tfrac{dy}{dx}$, $\tfrac{d^2 y}{dx^2}$, etc. We shall now proceed to find the formulas necessary for making such transformations.

Given $y =f(x)$, then from XXVI, (§ 33), we have

 (35) $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$, $\frac{dx}{dy} \ne 0$

giving $\tfrac{dy}{dx}$ in terms of $\tfrac{dx}{dy}$. Also, by XXV,(§ 33),

$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{dy}{dy} \left( \frac{dy}{dx} \right) \frac{dy}{dx}$

or

 (A) $\frac{d^2 y}{dx^2} = \frac{d}{dy} \left( \frac{1}{\frac{dx}{dy}} \right) \frac{dy}{dx}$.

But

$\frac{d}{dy} \left( \frac{1}{\frac{dx}{dy}} \right) = - \frac{\frac{d^2 x}{dy^2}}{\left( \frac{dx}{dy} \right)^2}$; and $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ from (35).

Substituting these in (A), we get

 (36) $\frac{d^2 y}{dx^2} = -\frac{ \frac{d^2 x}{dy^2} }{\left( \frac{dx}{dy} \right)^3}$,

giving $\tfrac{d^2 y}{dx^2}$ in terms of $\tfrac{dx}{dy}$ and $\tfrac{d^2 x}{dy^2}$. Similarly,

 (37) $\frac{d^3 y}{dx^3} = -\frac{ \frac{d^3 x}{dy^3} \frac{dx}{dy} - 3 \left( \frac{d^2 x}{dy^2} \right)^2 }{ \left( \frac{dx}{dy} \right) }$;

and so on for higher derivatives. This transformation is called changing the independent variable from x to y.

Illustrative Example 1. Change the independent variable from x to y in the equation

$3 \left( \frac{d^2 y}{dx^2} \right)^2 - \frac{dy}{dx} \frac{d^3 y}{dx^3} - \frac{d^2 y}{dx^2} \left( \frac{dy}{dx} \right)^2 = 0.$
Solution. Substituting from (35), (36), (37),
$3 \left( -\frac{ \frac{d^2 x}{dy^2} }{ \left( \frac{dx}{dy} \right )^3 } \right)^2 - \left( \frac{1}{ \frac{dx}{dy} } \right) \left( -\frac{ \frac{d^3 x}{dy^3} \frac{dx}{dy} - 3 \left( \frac{d^2 x}{dy^2} \right)^2 }{ \left( \frac{dx}{dy} \right)^5 } \right) - \left( -\frac{ \frac{d^2 x}{dy^2} }{ \left( \frac{dx}{dy} \right)^3 } \right) \left( \frac{1}{ \frac{dx}{dy} } \right)^2 = 0.$
Reducing, we get
$\frac{d^3 x}{dy^3} + \frac{d^2 x}{dy^2} = 0,$
a much simpler equation.

96. Change of the dependent variable. Let

 (A) $y = f(x),$

and suppose at the same time y is a function of z, say

 (B) $y = \psi(z).$

We may then express $\tfrac{dy}{dx}$, $\tfrac{d^2 y}{dx^2}$ etc., in terms of $\tfrac{dz}{dx}$, $\tfrac{d^2 z}{dx^2}$, etc., as follows

In general, z is a function of y by (B), §42; and since y is a function of x by (A), it is evident that z is a function of x. Hence by XXV we have

 (C) $\frac{dy}{dx} = \frac{dy}{dz} \frac{dz}{dx} = \psi'(z) \frac{dz}{dx}$. Also $\tfrac{d^2 y}{dx^2} = \tfrac{d}{dx} \left( \psi'(z) \tfrac{dz}{dx} \right) = \tfrac{dz}{dx} \tfrac{d}{dx} \psi'(z) + \psi'(z) \tfrac{d^2 z}{dx^2}$. By V, §33 But $\tfrac{d}{dx} \psi'(z) = \tfrac{d}{dz} \psi'(z) \tfrac{dz}{dx} = \psi''(z) \tfrac{dz}{dx}.$ By XXV, §33 (D) ∴ $\frac{d^2 y}{dx^2} = \psi''(z) \left( \frac{dz}{dx} \right)^2 + \psi'(z) \frac{d^2 z}{dx^2}.$

Similarly for higher derivatives. This transformation is called changing the dependent variable from y to z, the independent variable remaining x throughout. We will now illustrate this process by means of an example.

Illustrative Example 1. Having given the equation

 (E) $\frac{d^2 y}{dx^2} = 1 + \frac{2( 1 + y)}{1 + y^2} \left( \frac{dy}{dx} \right)^2,$

change the dependent variable from y to z by means of the relation

 (F) $y = \tan z.$

Solution. From (F),

$\frac{dy}{dx} = \sec^2 z \frac{dz}{dx}$, $\frac{d^2 y}{dx^2} = \sec^2 \frac{d^2 z}{dx^2} + 2 \sec^2 z \tan z \left( \frac{dz}{dx} \right)^2,$
Substituting in (E),
$\sec^2 z \frac{d^2 z}{dx^2} + 2 \sec^2 z \tan z \left( \frac{dz}{dx} \right)^2 = 1 \frac{2(1 + \tan z)}{1 + \tan^2 z} \left( \sec^2 z \frac{dz}{dx} \right)^2,$
and reducing, we get $\tfrac{d^2 z}{dx^2} - 2 \left( \tfrac{dz}{dx} \right)^2 = \cos^2 z$. Ans.

97. Change of the independent variable. Let y be a function of x, and at the same time let x (and hence also y) be a function of a new variable t. It is required to express

$\frac{dy}{dx},\ \frac{d^2 y}{dx^2}$, etc.,

in terms of new derivatives having t as the independent variable.

By XXV, §33,

 $\frac{dy}{dt}$ $= \frac{dy}{dx} \frac{dx}{dt}$, or (A) $\frac{dy}{dx}$ $= \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }$. Also $\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right)$ $= \frac{d}{dt} \left( \frac{dy}{dx} \right) \frac{dt}{dx} = \frac{ \frac{d}{dt} \left( \frac{dy}{dx} \right) }{ \frac{dx}{dt} }$ But differentiating (A) with respect to t, $\frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } \right)$ $= \frac{ \frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^2 }$

Therefore

 (B) $\frac{d^2 y}{dx^2}= \frac{ \frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^3}$;

and so on for higher derivatives. This transformation is called changing the independent variable from x to t. It is usually better to work out examples by the methods illustrated above rather than by using the formulas deduced.

Illustrative Example 1. Change the independent variable from x to t in the equation.

 (C) $x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + y = 0$ (D) $\ x = e^t\$.
Solution.
 $\frac{dx}{dt} = e^t\$, therefore (E) $\frac{dt}{dx}= e^{-t}\$. Also $\frac{dy}{dx}= \frac{dy}{dt} \frac{dt}{dx}$; therefore (F) $\frac{dy}{dx} = e^{-t} \frac{dy}{dt}$ Also $\frac{d^2 y}{dx^2} = e^{-t} \frac{d}{dx} \left( \frac{dy}{dt} \right) - \frac{dy}{dt} e^{-t} \frac{dt}{dx} = e^{-t} \frac{d}{dt} \left( \frac{dy}{dt} \right) \frac{dt}{dx} - \frac{dy}{dt} e^{-t} \frac{dt}{dx}.$
Substituting in the last result from (E),
 (G) $\frac{d^2 y}{dx^2}= e^{-2t} \frac{d^2 y}{dt^2} - \frac{dy}{dt} e^{-2t};$
Substituting (D), (F), (G) in (C),
$e^{2t} \left( e^{-2t} \frac{d^2 y}{dt^2} - \frac{dy}{dt} e^{-2t} \right) + e^t \left( e^{-t} \frac{dy}{dt} \right) + y = 0;$
and reducing, we get
$\frac{d^2 y}{dt^2} + y = 0\$ Ans.

Since the formulas deduced in the Differential Calculus generally involve derivatives of y with respect to x, such formulas as (A) and (B) are especially useful when the parametric equations of a curve are given. Such examples were given in §66, and many others will be employed in what follows.

98. Simultaneous change of both independent and dependent variables. It is often desirable to change both variables simultaneously. An important case is that arising in the transformation from rectangular to polar coördinates. Since

$x = \rho \cos \theta$ and $y = \rho \sin \theta,$

the equation

$f( x, y ) = 0$

becomes by substitution an equation between ρ and θ, defining ρ as a function of θ. Hence ρ, x, y are all functions of θ.

Illustrative Example 1. Transform the formula for the radius of curvature (42), §103,

 (A) $R= \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} }{ \frac{d^2 y}{dx^2} },$

into polar coördinates.

Solution. Since in (A) and (B), §97, t is any variable on which x and y depend, we may in this case let $t = \theta$, giving
 (B) $\frac{dy}{dx} = \frac{ \frac{dy}{d\theta} }{ \frac{dx}{d\theta} }$, and (C) $\frac{d^2 y}{dx^2} = \frac{ \frac{dx}{d\theta} \frac{d^2 y}{d\theta^2} - \frac{dy}{d\theta} \frac{d^2 x}{d\theta^2} }{ \left( \frac{dx}{d\theta} \right)^3 }$
Substituting (B) and (C) in (A), we get
 $R = \left[ \frac{ \left( \frac{dx}{d\theta} \right)^2 + \left( \frac{dy}{d\theta} \right)^2 }{ \left( \frac{dx}{d\theta} \right)^2 } \right]^{\frac{3}{2}} \div \frac{ \frac{dx}{d\theta} \frac{d^2 y}{d\theta^2} - \frac{dy}{d\theta} \frac{d^2 x}{d\theta^2} }{ \left( \frac{dx}{d\theta} \right)^3 }$, or (D) $R = \frac{ \left[ \left( \frac{dx}{d\theta} \right)^2 + \left( \frac{dy}{d\theta} \right)^2 \right]^{\frac{3}{2}} }{ \frac{dx}{d\theta} \frac{d^2 y}{d\theta^2} - \frac{dy}{d\theta} \frac{d^2 x}{d\theta^2} }$.
But since $x = \rho \cos \theta$ and $y = \rho \sin \theta$, we have
$\frac{dx}{d\theta} = - \rho \sin \theta + \cos \theta \frac{d\rho}{d\theta}$; $\frac{dy}{d\theta} = \rho \cos \theta + \sin \theta \frac{d\rho}{d\theta};$
$\frac{d^2 x}{d\theta^2} = -\rho \cos \theta - 2 \cos \theta \frac{d\rho}{d\theta} + \cos \theta \frac{d^2 \rho}{d\theta^2}$; $\frac{d^2 y}{d\theta^2} = -\rho \sin \theta + 2 \cos \theta \frac{d\rho}{d\theta} + \sin \theta \frac{d^2 \rho}{d\theta^2}.$
Substituting these in (D) and reducing,
$R = \frac{ \left[ \rho^2 + \left( \frac{d\rho}{d\theta} \right)^2 \right]^{\frac{3}{2}} }{ \rho^2 2 \left( \frac{d\rho}{d\theta} \right)^2 - \rho \frac{d^2 \rho}{d\theta^2} }$. Ans.
EXAMPLES

Change the independent variable from x to y in the four following equations:

 1. $R = \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} }{ \frac{d^2 y}{dx^2} }$ Ans. $R = -\frac{ \left[ 1 + \left( \frac{dx}{dy} \right)^2 \right]^{\frac{3}{2}} }{ \frac{d^2 x}{dy^2} }$ 2. $\frac{d^2 y}{dx^2} + 2 y \left( \frac{dy}{dx} \right)^2 = 0$. $\frac{d^2 x}{dy^2} - 2 y \frac{dx}{dy} = 0$. 3. $x \frac{d^2 y}{dx^2} + \left( \frac{dy}{dx} \right)^3 - \frac{dy}{dx} = 0$ $x \frac{d^2 y}{dx^2} - 1 + \left( \frac{dx}{dy} \right)^2 = 0$. 4. $\left( 3 a \frac{dy}{dx} + 2 \right) \left( \frac{d^2 y}{dx^2} \right)^2 = \left( a \frac{dy}{dx} + 1 \right) \frac{dy}{dx} \frac{d^3 y}{dx^3}$ $\left( \frac{d^2 x}{dy^2} \right)^2 = \left( \frac{dx}{dy} + a \right) \frac{d^3 x}{dy^3}$.

Change the dependent variable from y to z in the following equations:

 5. $(1 + y)^2 \left( \frac{d^3 y}{dx^3} - 2 y \right) + \left( \frac{dy}{dx} \right)^2 = 2 \left( 1 + y \right) \frac{dy}{dx} \frac{d^2 y}{dx^2}, y = z^2 + 2 z$. Ans. $(z + 1) \frac{d^3 x}{dx^3} = \frac{dz}{dx} \frac{d^2 z}{dx^2} + z^2 + 2 z$. 6. $\frac{d^2 y}{dx^2} = 1 + \frac{2 (1 + y)}{1 + y^2} \left( \frac{dy}{dx} \right)^2, y = \tan z$. Ans. $\frac{d^2 z}{dx^2} - 2 \left( \frac{dz}{dx} \right)^2 = \cos^2 z$. $y^2 \frac{d^3 y}{dx^3} - \left( 3 y \frac{dy}{dx} + 2 x y^2 \right) \frac{d^2 y}{dx^2} + \left\{ 2 \left( \frac{dy}{dx} \right)^2 2xy \frac{dy}{dx} + 3x^2 y^2 \right\} \frac{dy}{dx} + x^3 y^3 = 0, y = e^z$. Ans. $\frac{d^3 z}{dx^3} - 2 x \frac{d^2 z}{dx^2} + 3 x^2 \frac{dz}{dx} + x^3 = 0$.

Change the independent variable in the following eight equations:

 8. $\frac{d^2 y}{dx^2} - \frac{x}{1 - x^2} \frac{dy}{dx} + \frac{y}{1 - x^2} = 0$, $x = \cos t$. Ans. $\frac{d^2 y}{dt^2} + y = 0$. 9. $(1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0$, $x = \cos z$. $\frac{d^2 y}{dz^2} = 0$. 10. $(1 - y^2) \frac{d^2 u}{dy^2} - y \frac{du}{dy} + a^2 u = 0$, $y = \sin x$. $\frac{d^2 u}{dx^2} + a^2 u = 0$. 11. $x^2 \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} + \frac{a^2}{x^2} y = 0$, $x = \frac{1}{z}$. $\frac{d^2 y}{dz^2} + a^2 y = 0$. 12. $x^3 \frac{d^3 v}{dx^3} + 3x^2 \frac{d^2 v}{dx^2} + x \frac{dv}{dx} + v = 0$, $x = e^t$. $\frac{d^3 v}{dx^3} + v = 0$. 13. $\frac{d^2 y}{dx^2} + \frac{2x}{1 + x^2} \frac{dy}{dx} + \frac{y}{(1 + x^2)^2} = 0$, $x = \tan \theta$. $\frac{d^2 y}{d\theta^2} + y = 0$. 14. $\frac{d^2 u}{ds^2} + su \frac{du}{ds} + \sec^2 s = 0$, $s = \arctan t$. Ans. $(1 + t^2) \frac{d^2 u}{dt^2} + (2t + x \arctan t) \frac{du}{dt} + 1 = 0$. 15. $x^4 \frac{d^2 y}{dx^2} + a^2 y = 0$, $x = \frac{1}{z}$. Ans. $\frac{d^2 y}{dz^2} + \frac{2}{z} \frac{dy}{dz} + a^2 y = 0$.

In the following seven examples the equations are given in parametric form.

Find $\tfrac{dy}{dx}$ and $\tfrac{d^2 y}{dx^2}$ in each case:

 16. $x = 7 + t^2, y = 3 + t^2 - 3t^4$. Ans. $\frac{dy}{dx} = 1 - 6 t^2, \frac{d^2 y}{dx^2} = -6$. 17. $x = \cot t, y = \sin^3 t$. Ans. $\frac{dy}{dx} = -3 \sin^4 t \cos t, \frac{d^2 y}{dx^2} = 3 \sin^5 t (4 - 5 \sin^2 t)$. 18. $x = a(\cos t + \sin t), y = a(\sin t - t \cos t)$. Ans. $\frac{dy}{dx} = \tan t, \frac{d^2 y}{dx^2} = \frac{1}{at \cos^3 t}$. 19. $x = \frac{1 - t}{1 + t}, y = \frac{2t}{1 + t}$. 20. $x = 2 t, y = 2 - t^2$. 21. $x = 1 - t^2, y = t^3$. 22. $x = a \cos t, y = b \sin t$. 23. Transform $\frac{ x \frac{dy}{dx} - y}{ \sqrt{1 + \left( \frac{dy}{dx} \right)^2} }$ by assuming $x = \rho \cos \theta, y = \rho \sin \theta$. Ans. $\frac{\rho^2}{ \sqrt{ \rho \left( \frac{d\rho}{d\theta} \right)^2 } }$. 24. Let $f(x, y) = 0$ be the equation of a curve. Find an expression for its slope $\left( \frac{dy}{dx} \right)$ in terms of polar coördinates. Ans. $\frac{dy}{dx} = \frac{ \rho \cos \theta + \sin \theta \frac{d\rho}{d\theta} }{ -\rho \sin \theta + \cos \theta \frac{d\rho}{d\theta} }$.