Tutorial: Calculus
Francis Wright, June 2018
Click on a problem header to show/hide my REDUCE solution.
Find the set of all \(x \in \mathbb{R}\) which satisfy the inequality
\[ |x - 1| \le x^2 - 1. \]
plot({abs(x-1), x^2-1}, x=(-3 .. 3)); % Clearly, the problem is equivalent to -(x-1) <= x^2-1. solve(-(x-1)=x^2-1); % Hence the solution is x <= -2 or x >= 1.
Let \[ f(x) = \frac{x^2 - 4x + 4}{x^2 - x - 2} \]
for all \(x \in \mathbb{R} \setminus \{-1, 2\}\).
Determine whether each of the following limits exist,
giving the value if it exists:
\[ \lim_{x \to -1} f(x); \quad \lim_{x \to 2} f(x); \quad \lim_{x \to \infty} f(x). \]
Determine whether \(f\) has a continuous extension at \(x = -1\) or \(x = 2\),
defining the extension if it exists.
f := (x^2-4x+4)/(x^2-x-2); limit(f, x, -1); % i.e. no finite limit limit(f, x, 2); limit(f, x, infinity); % Note that on gcd; f; off gcd; % Hence no continuous extension at x = -1 but a continuous extension as above at x = 2.
Determine the derivative of the function arctan.
df(atan x, x);
Find the equation of the tangent to the curve
\(y^3 + x^2y - 3x + 1 = 0\) at the point \((1, 1)\).
depend y, x; curve := y^3 + x^2*y - 3x + 1; % = 0 df(curve, x); solve(ws, df(y,x)); m := sub(x=1, y=1, rhs first ws); tangent := (y-1) - m*(x-1); % = 0 % The following trick plots the implicitly-defined curve and tangent line % together, but unfortunately with a gap and with the x-axis vertical: plot(curve*tangent=0, x=(0 .. 2), y=(0 .. 2), points=100); nodepend y, x;
Find the area of the region bounded above by the curve
\(x^2 + y^2 = 2\) and below by the curve \(y = x^2\).
solve(x^2 + y^2 = 2, y); above := rhs first ws; below := x^2; solve(above = below); int(above - below, x, -1, 1);
Evaluate \[ \int (x + 2) \ln(x - 3) \, dx . \]
define ln = log; % or just use log instead of ln int((x+2)*ln(x-3), x); % This form is slightly tidier: factor log; ws;
Consider the curve \(y = f(x)\) for the function \(f(x) = 3x^\frac23(5 - x)\).
- Identify the domain of \(f\) and determine whether or not \(f\) is an even function or an odd function.
- Find \(f'(x)\) and \(f''(x)\).
- Find the critical points of \(f\), determine where \(f\) is increasing or decreasing, and determine the behavior of \(f\) at each of its critical points.
- Find the inflexion points for \(f\), if any occur, and determine the concavity of the curve.
- Determine the behavior of \(f(x)\) as \(x \to \pm\infty\) and identify any asymptotes.
- Plot key points, such as intercepts, critical points, and points of inflexion, and sketch the curve.
operator f; let f(~x) => 3x^(2/3)*(5-x); % Domain is x >= 0 since x^(2/3) is undefined for x < 0. % Hence f(x) is neither even nor odd. Also, formally: if f(x) = f(-x) then even else if f(x) = -f(-x) then odd else neither; df(f(x), x); operator fp; for all x saveas fp(x); df(f(x), x, 2); operator fpp; for all x saveas fpp(x); solve fp(x); % One critical point at x = 2. fpp(rhs first ws); % This critical point is a maximum. Hence f is increasing for x < 2 and decreasing for x > 2. solve fpp(x); % No inflexion points (within the domain of f). % Hence (since f'' < 0) f is (strictly) concave (downwards). limit(f(x), x, infinity); plot(f(x), x=(0 .. 6));
Consider the function \[ f(x, y) = \frac{2x - xy - 3y + 6}{y - 2}, y \not= 2. \]
Find the limit of \(f\) as \((x, y) \to (1, 2)\).
f := (2x - x*y -3y + 6)/(y - 2); sub(x=1, y=2, f);
Find the directional derivative of the function
\[f(x, y, z) = xy + xz + yz\]
at the point \((-1, 1, -2)\) in the direction of the vector
\(\mathbf{A} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\).
f := x*y + x*z + y*z; load_package avector; g := grad f; x := -1$ y := 1$ z := -2$ g := avec(g(0),g(1),g(2)); clear x,y,z; A := avec(1, -2, 2); % i - 2j + 2k g dot (A/vmod A);
Find the linearisation of \[f(x, y) = 7x + 3y - 11\]
at the point \((2, 3)\).
f := 7x + 3y - 11; {df(f,x), df(f,y), f}; sub(x=2, y=3, ws); x*first ws + y*second ws + third ws;
Obtain the limit as \(n \to \infty\) for the sequence
\[ a_n = \left(1 + \frac5n\right)^n. \]
an := (1 + 5/n)^n; limit(an, n, infinity);
Find the sum of the series \[ \sum_{k=0}^\infty (-2)^k \frac3{4^k}. \]
% Must do some of the work by hand for this example! % s := (-2)^k*3/4^k; % doesn't simplify (even with on combineexpt), so s := (-2/4)^k*3; sum(s, k, 0, infinity); % obviously should evaluate to 2. % sum(s, k, 0, n); limit(ws, n, infinity); % doesn't simplify
Sketch the region of integration and evaluate the integral
\[ \int_0^1 \int_y^\sqrt y (3x - 1) \, dx \, dy. \]
Then reverse the order of integration and re-evaluate the integral.
plot({x,x^2}, x=(0 .. 1)); % As in the question: int( int(3x-1, x, y, sqrt y), y, 0, 1); % With the order of integration reversed: int( int(3x-1, y, x^2, x), x, 0, 1);
Find the Jacobian \(\partial(x,y,z)/\partial(u,v,w)\) for the
transformation \(x = v + 2w, y = u + vw, z = uv^2.\)
load_package linalg; mat_jacobian({v+2w, u+v*w, u*v^2}, {u, v, w});
Solve the differential equation \[ \frac{dy}{dx} = e^{y-x} \]
and confirm the solution.
depend y, x; df(y,x) = exp(y-x); saveas ode; odesolve ode; solve(ws, y); sub(ws, ode); % This equation is clearly true, even if REDUCE % doesn't seem to know that exp(2 arbint i pi) = 1. nodepend y, x;
Let \[ f(x,y) = \frac{x^2}4 + y^2. \]
Sketch the level curve \(f = 2\).
Calculate the gradient vector \(\nabla f\) at the point
\((2, -1)\) and sketch it. Then calculate the equation for the
tangent line to the level curve at the same point and sketch it.
f := x^2/4 + y^2; plot(f - 2 = 0, x = (-3 .. 3), y = (-2 .. 2)); load_package avector; g := grad f; x := 2$ y := -1$ p := avec(x,y,0); g := avec(g(0), g(1), 0); clear x, y; % There is currently no way to plot this vector together with the curve. tangent := avec(-g(1), g(0), 0); % A general point on the tangent line is given for any real alpha by line := p + alpha*tangent; eqn := avec(x,y,0) - line; solve(eqn(0), alpha); sub(ws, eqn(1)); % The Cartesian equation of the line is first solve(ws, y); y_tangent := rhs ws$ % Until plot handles mixed data better... eps := 1e-6$ solve(f = 2, y)$ plot({rhs first ws, rhs second ws, y_tangent}, x = (-2 sqrt 2+eps .. 2 sqrt 2-eps), y = (-sqrt 2+eps .. sqrt 2-eps));
Use the method of Lagrange multipliers to find the extreme points of
the function \[ f(x,y) = xy \]
subject to the condition \[ x^2 + y^2 = 9. \]
Determine the value of the function \(f\) at each of the extreme points.
f := x*y; g := x^2 + y^2 - 9; % = 0 % Define a Lagrangian function L with Lagrange multiplier lam and find its critical points: L := f - lam*g; {df(L,x), df(L,y), g}; solns := solve ws; for each soln in solns do write append(rest soln, sub(soln, {f, g})); % Hence x = y = +/- 3/sqrt 2 are maxima with f = 9/2 and x = -y = +/- 3/sqrt 2 are minima with f = -9/2.
Find the Taylor series generated by \(f(x) = e^{2x}\) at \(x = 0\).
f := exp(2x); df(f,x,1); df(f,x,2); df(f,x,3); % Conclude that df(f,x,n) evaluated at x = 0 is sub(x=0, 2^n*exp(2x)); % Hence the nth term is operator term; let term ~n => 2^n*x^n/factorial n; % The Taylor series is the sum of this general term for n from 0 to infinity. % The Taylor polynomial to degree 6 is for n := 0:6 sum term n; % This looks better like this: on div, revpri; ws; off div, revpri; % This is the same as load_package "taylor"; taylorprintterms := all$ taylor(f, x, 0, 6);