2015-11-02-225332.sagews

1

Use Newton's Method with 5 iterations

f(x)=4*x^3-9*x^2+10
plot(f,xmin=-2,xmax=0,ymin=-5,ymax=5)
df(x)=derivative(f,x)
r=-1
n=5
for i in range(n):
    r=r-f(r)/df(r)
    N(r)

-0.900000000000000 -0.892052469135802 -0.892003706245178 -0.892003704415253 -0.892003704415253

2

Find $dy/dx$ when $x^2+xy+y^2= x^2y^2$

%var y
y=function('y',x)
solve(derivative(x^3-3*y^2==4*x+2*y,x),derivative(y,x))
show(_)

Error in lines 3-3 Traceback (most recent call last): File "/projects/sage/sage-6.10/local/lib/python2.7/site-packages/smc_sagews/sage_server.py", line 905, in execute exec compile(block+'\n', '', 'single') in namespace, locals File "", line 1, in <module> File "/projects/sage/sage-6.10/local/lib/python2.7/site-packages/sage/calculus/functional.py", line 130, in derivative return f.derivative(*args, **kwds) File "sage/symbolic/expression.pyx", line 3765, in sage.symbolic.expression.Expression.derivative (/projects/sage/sage-6.10/src/build/cythonized/sage/symbolic/expression.cpp:22854) return multi_derivative(self, args) File "sage/misc/derivative.pyx", line 222, in sage.misc.derivative.multi_derivative (/projects/sage/sage-6.10/src/build/cythonized/sage/misc/derivative.c:2861) F = F._derivative(arg) File "sage/symbolic/expression.pyx", line 3833, in sage.symbolic.expression.Expression._derivative (/projects/sage/sage-6.10/src/build/cythonized/sage/symbolic/expression.cpp:23333) raise TypeError("argument symb must be a symbol") TypeError: argument symb must be a symbol

3

Suppose that x, y, and z are all functions of t, and $x^3+y^3=z^3$

%var t
x=function('x',t)
y=function('y',t)
z=function('z',t)
solve(derivative(x+y==2*x*y*z,t),derivative(z,t))
show(_)
solve(derivative(x+y==2*x*y*z,t),derivative(y,t))
show(_)
solve(derivative(x+y==2*x*y*z,t),derivative(x,t))
show(_)

[D[0](z)(t) == -1/2*((2*y(t)*z(t) - 1)*D[0](x)(t) + (2*x(t)*z(t) - 1)*D[0](y)(t))/(x(t)*y(t))]

[$\displaystyle D[0]\left(z\right)\left(t\right) = -\frac{{\left(2 \, y\left(t\right) z\left(t\right) - 1\right)} D[0]\left(x\right)\left(t\right) + {\left(2 \, x\left(t\right) z\left(t\right) - 1\right)} D[0]\left(y\right)\left(t\right)}{2 \, x\left(t\right) y\left(t\right)}$]

[D[0](y)(t) == -(2*x(t)*y(t)*D[0](z)(t) + (2*y(t)*z(t) - 1)*D[0](x)(t))/(2*x(t)*z(t) - 1)]

[$\displaystyle D[0]\left(y\right)\left(t\right) = -\frac{2 \, x\left(t\right) y\left(t\right) D[0]\left(z\right)\left(t\right) + {\left(2 \, y\left(t\right) z\left(t\right) - 1\right)} D[0]\left(x\right)\left(t\right)}{2 \, x\left(t\right) z\left(t\right) - 1}$]

[D[0](x)(t) == -(2*x(t)*y(t)*D[0](z)(t) + (2*x(t)*z(t) - 1)*D[0](y)(t))/(2*y(t)*z(t) - 1)]

[$\displaystyle D[0]\left(x\right)\left(t\right) = -\frac{2 \, x\left(t\right) y\left(t\right) D[0]\left(z\right)\left(t\right) + {\left(2 \, x\left(t\right) z\left(t\right) - 1\right)} D[0]\left(y\right)\left(t\right)}{2 \, y\left(t\right) z\left(t\right) - 1}$]