Exercises

Change the independent variable from $x$ to $y$ in the 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. $\frac{d^2 y}{dx^2} + 2 y \left( \frac{dy}{dx} \right)^2 = 0$.

   Ans. $\frac{d^2 x}{dy^2} - 2 y \frac{dx}{dy} = 0$.

3. $x \frac{d^2 y}{dx^2} + \left( \frac{dy}{dx} \right)^3 - \frac{dy}{dx} = 0$.

   Ans. $x \frac{d^2 y}{dx^2} - 1 + \left( \frac{dx}{dy} \right)^2 = 0$.

4. $\left( 3 a \frac{dy}{dx} + 2 \right) \left( \frac{d^2 y}{dx^2} \right)^2 = \left( a \frac{dy}{dx} + 1 \right) \frac{dy}{dx} \frac{d^3 y}{dx^3}$.

   Ans. $\left( \frac{d^2 x}{dy^2} \right)^2 = \left( \frac{dx}{dy} + a \right) \frac{d^3 x}{dy^3}$.

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

1. $(1 + y)^2 \left( \frac{d^3 y}{dx^3} - 2 y \right) + \left( \frac{dy}{dx} \right)^2 = 2 \left( 1 + y \right) \frac{dy}{dx} \frac{d^2 y}{dx^2}, y = z^2 + 2 z$.

   Ans. $(z + 1) \frac{d^3 x}{dx^3} = \frac{dz}{dx} \frac{d^2 z}{dx^2} + z^2 + 2 z$.

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

   Ans. $\frac{d^2 z}{dx^2} - 2 \left( \frac{dz}{dx} \right)^2 = \cos^2 z$.

3. $y^2 \frac{d^3 y}{dx^3} - \left( 3 y \frac{dy}{dx} + 2 x y^2 \right) \frac{d^2... ...)^2 2xy \frac{dy}{dx} + 3x^2 y^2 \right\} \frac{dy}{dx} + x^3 y^3 = 0, y = e^z$.

   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:

1. $\frac{d^2 y}{dx^2} - \frac{x}{1 - x^2} \frac{dy}{dx} + \frac{y}{1 - x^2} = 0$, $x = \cos\, t$.

   Ans. $\frac{d^2 y}{dt^2} + y = 0$.

2. $(1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0$, $x = \cos\, z$.

   Ans. $\frac{d^2 y}{dz^2} = 0$.

3. $(1 - y^2) \frac{d^2 u}{dy^2} - y \frac{du}{dy} + a^2 u = 0$, $y = \sin\, x$.

   Ans. $\frac{d^2 u}{dx^2} + a^2 u = 0$.

4. $x^2 \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} + \frac{a^2}{x^2} y = 0$, $x = \frac{1}{z}$.

   Ans. $\frac{d^2 y}{dz^2} + a^2 y = 0$.

5. $x^3 \frac{d^3 v}{dx^3} + 3x^2 \frac{d^2 v}{dx^2} + x \frac{dv}{dx} + v = 0$, $x = e^t$.

   Ans. $\frac{d^3 v}{dx^3} + v = 0$.

6. $\frac{d^2 y}{dx^2} + \frac{2x}{1 + x^2} \frac{dy}{dx} + \frac{y}{(1 + x^2)^2} = 0$, $x = \tan\theta$.

   Ans. $\frac{d^2 y}{d\theta^2} + y = 0$.

7. $\frac{d^2 u}{ds^2} + su \frac{du}{ds} + \sec^2 s = 0$.

   Ans. $s = \arctan\, t$.

8. $x^4 \frac{d^2 y}{dx^2} + a^2 y = 0$, $x = \frac{1}{z}$.

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

1. $x=7+t^2$, $y =3 + t^2 - 3t^4$.

   Ans. $\frac{dy}{dx} = 1 - 6 t^2$, $\frac{d^2 y}{dx^2} = -6$.

   We shall solve this using SAGE.

   [fontsize=\scriptsize,fontfamily=courier,fontshape=tt,frame=single,label=\sage]
   
   sage: t = var("t")
   sage: x = 7 + t^2
   sage: y = 3 + t^2 - 3*t^4
   sage: f = (x, y)
   sage: p = parametric_plot(f, 0, 1)
   sage: D_x_of_y = diff(y,t)/diff(x,t); D_x_of_y
   (2*t - 12*t^3)/(2*t)
   sage: solve(D_x_of_y == 0,t)
   [t == -1/sqrt(6), t == 1/sqrt(6)]
   sage: t0 = solve(D_x_of_y == 0,t)[1].rhs()
   sage: (x(t0),y(t0))
   (43/6, 37/12)
   sage: D_xx_of_y = (diff(y,t,t)*diff(x,t)-diff(x,t,t)*diff(y,t))/diff(x,t)^2; D_xx_of_y
   (2*t*(2 - 36*t^2) - 2*(2*t - 12*t^3))/(4*t^2)
   sage: D_xx_of_y(t0)
   -12/sqrt(6)
   

   This tells us that the critical point is at $(43/6, 37/12)=(7.166.., 3.0833..)$, which is a maximum. The plot in Figure 11.1 illustrates this.

   Figure 11.1: Plot for Exercise 11.5-16, $x=7+t^2$, $y =3 + t^2 - 3t^4$.

   

2. $x = \cot\, t$, $y = \sin^3t$.

   Ans. $\frac{dy}{dx} = -3 \sin^4 t \cos t$, $\frac{d^2 y}{dx^2} = 3 \sin^5 t (4 - 5 \sin^2 t)$.

3. $x = a(\cos\, t + \sin\, t)$, $y = a(\sin\, t - t\cos\, t)$.

   Ans. $\frac{dy}{dx} = \tan t$, $\frac{d^2 y}{dx^2} = \frac{1}{at \cos^3 t}$.

4. $x = \frac{1 - t}{1 + t}$, $y = \frac{2t}{1 + t}$.

5. $x = 2t$, $y = 2 - t^2$.

6. $x = 1 - t^2$, $y = t^3$.

7. $x = a\cos\, t$, $y = b\sin\, t$.

8. 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. $\frac{\rho^2}{ \sqrt{ \rho \left( \frac{d\rho}{d\theta} \right)^2 } }$.

9. 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 coordinates.

   Ans. $\frac{dy}{dx} = \frac{ \rho \cos \theta + \sin \theta \frac{d\rho}{d\theta} }{ -\rho \sin \theta + \cos \theta \frac{d\rho}{d\theta} }$.