vectorialcalculus (version 1.0.5)

related_rates_grad: Related rates via the gradient (implicit constraint)

Description

Computes related rates for an implicit constraint using the gradient. Let $$ g(\mathbf{x}) = 0,\quad \mathbf{x}=\mathbf{x}(t)\in\mathbb{R}^k. $$ Differentiating with respect to time yields $$ \nabla g(\mathbf{x}(t))\cdot \mathbf{x}'(t) = 0. $$ If all components of \(\mathbf{x}'(t)\) are known except one, the missing rate is determined by this orthogonality condition (velocity tangent to the level set).

Usage

related_rates_grad(g, x, known_rates, solve_for, var_names = NULL, h = 1e-06)

Value

A list with components:

rate

Numeric scalar. The solved rate (the requested component of x'(t)).

rates

Named numeric vector length k. Full rate vector x'(t). Names are paste0("d", var_names).

grad

Numeric vector length k. Gradient of g at x.

dot

Numeric scalar. Dot product grad . rates (should be near 0).

gx

Numeric scalar. g(x) value (should be near 0 if x is on the constraint).

Arguments

g

Function. A scalar function g(x1, x2, ..., xk) defining the implicit constraint g = 0. It must accept k numeric arguments and return a numeric scalar.

x

Numeric vector of length k. Point where rates are evaluated. This point should satisfy g(x) = 0 (approximately).

known_rates

Numeric vector of known rates. It can be: (1) a named numeric vector with names matching var_names (e.g. "x","y"), (2) a named numeric vector with names matching paste0("d", var_names) (e.g. "dx","dy"), or (3) an unnamed numeric vector of length k with NA for the unknown component.

solve_for

Integer or character. Which rate to solve for. If integer, it is the position in var_names (1..k). If character, it may be either a variable name (e.g. "y") or a rate name (e.g. "dy").

var_names

Character vector of length k. Variable names. If NULL, defaults to c("x1","x2",...,"xk").

h

Numeric scalar. Step size for numerical partial derivatives.

Examples

# Ladder (implicit circle): x^2 + y^2 = L^2
# Suppose L = 5, at the instant (x,y) = (4,3) and dx/dt = -1.
# Then dy/dt = -(x/y) dx/dt = (4/3).
g <- function(x, y) x^2 + y^2 - 25

out <- related_rates_grad(
  g = g,
  x = c(4, 3),
  known_rates = c(dx = -1, dy = NA),
  solve_for = "dy",
  var_names = c("x", "y")
)
out$rate
out$rates