GAC016 Matemática III: Cálculo
Matemática do GAC Tópico 3 11:30 Narração em inglês · Legendas em inglês + 中文 gravadas
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An average speed does not tell you every instantaneous speed, and a signed displacement does not always equal total distance. Calculus makes those distinctions precise. GAC zero one six develops differentiation, integration and their applications, and this lesson works the core method in each area: state what is known, write the mathematics in symbols, substitute, and interpret the result inside the stated model.
Before any calculation, state the independent variable, the domain and the units. The notation d y over d x is a derivative with respect to x, while a dot over y normally denotes a derivative with respect to time. Those notations describe the same operation only when their independent variables agree, so notation is part of the data. For applications, describe what a rate or accumulated value means in the stated model. A negative volume-change rate of three cubic centimetres per second means the volume is falling at that rate, rather than that the volume itself is negative. And your centre's current brief determines the actual assessment; these practice sheets do not establish a university credit decision or an official examination pattern.
A derivative is a rate of change: how fast y changes as x changes. Geometrically it is the gradient of the tangent at a point. The definition is a limit: f prime of x equals the limit, as h goes to zero, of, open f of x plus h, minus f of x, close, over h. Differentiation from first principles uses that difference quotient for nonzero h and then takes its limit. The definition also tells you when a derivative fails: unequal one-sided limits, as for the absolute value of x at zero, prevent a derivative even when the function is continuous.
The power rule says that y equals x to the n differentiates to n x to the n minus one, on a domain where the power is differentiable. The product rule differentiates u v as u prime v plus u v prime. The quotient rule gives u prime v minus u v prime, over v squared, where v is nonzero. The chain rule says that a composition y equals f of g of x differentiates to f prime of g of x, times g prime of x. Match the rule to the structure of the expression before touching the algebra, because each rule carries its own condition, and the condition is where marks quietly live.
Stationary points occur where f prime of x equals zero, which is where the tangent is flat. The second derivative then classifies a stationary point when it is nonzero: a negative second derivative gives a local maximum, and a positive one gives a local minimum. A zero second derivative is inconclusive, and you must fall back on derivative signs on either side, or other evidence. And a stationary inflection can be neither an extremum: the derivative is zero, the curve pauses, and no maximum or minimum occurs. Classification is a conclusion from evidence, not a reflex to the equals-zero sign.
The curve y equals x squared has a tangent at the point one, one, and a dashed secant joining one, one to two, four. The secant slope is the average rate: four minus one, over two minus one, which is three. The derivative comes from first principles: the difference quotient of x squared simplifies to two x plus h, whose limit is two x. So the tangent slope at x equal to one is two times one, which is two. The average slope over a finite interval need not equal the instantaneous slope at either endpoint; the secant measures the journey, the tangent measures the moment. Use practice sheet three point one to test the method, domain and interpretation against its solutions.
Pause and choose. The derivative of x squared is two x, so at x equal to three the tangent slope is six. Nine is the function's value, which answers how high, not how steep. Five is the average slope between one and three, a secant question. And three is the location, not the slope. The discipline is the order: differentiate, then substitute, and state which question the number answers.
Indefinite integration finds antiderivatives: a family of functions differing by a constant, which is why the plus c is not optional. The power rule integrates x to the n as x to the n plus one over n plus one, plus c, for n not equal to minus one. Definite integration measures signed accumulation between limits and gives a number, F of b minus F of a, by the fundamental theorem of calculus for a continuous integrand. Integration by substitution reverses the chain rule. Signed accumulation can differ from geometric area, which is the next frame's whole story.
Here is the warning stated plainly, because it costs more marks than any calculus error does. Forgetting plus c on an indefinite integral is the single most frequent lost mark in the module, and it is lost on questions you have otherwise answered correctly. The indefinite integral names a whole family of antiderivatives, every vertical shift of the one you found, and the constant is how the answer says so. Write the plus c in the same pen stroke as the integral, not as a repair after checking the solution.
The line y equals x crosses the axis at zero, with shaded regions on both sides over minus one to one. The definite integral is the signed accumulation: the antiderivative x squared over two, evaluated from minus one to one, is one half minus one half, which is zero. The negative and positive contributions cancel. Geometric area instead adds the magnitudes of the two triangular regions: minus the integral from minus one to zero, plus the integral from zero to one, which is one half plus one half, one. Same curve, same interval, two different questions, two different answers. Ask which one the task means before integrating. Use practice sheet three point two to test the method, domain and interpretation against its solutions.
Pause and choose. The antiderivative of x is x squared over two. Evaluating at three gives nine halves, and at zero gives zero, so the signed integral is nine over two. Cancellation belongs to intervals crossing the axis, like the worked example's minus one to one. Nine is the function's value at the endpoint, and three is just the length of the interval. Where the interval lies decides whether the signed answer and the geometric answer agree; here they do.
Optimisation finds the largest or smallest value of a quantity, and it is a method rather than a guess. Write the quantity as a function of one variable on a stated feasible domain. Differentiate and find the stationary candidates. Then justify the required optimum, because a stationary point is only a candidate until classified. Check included boundaries, where a largest value can sit with no flat tangent at all, and integer constraints, where the true best integer point may neighbour the stationary one. The justification is the marks; the derivative alone is the easy half.
Related rates chain together: d V over d t equals d V over d r, times d r over d t, exactly when the variables are linked in the stated model. The area between two curves is the integral of f minus g, from a to b, where f is the upper curve on that whole interval; if the curves cross, split the integral at each crossing, because the upper curve changes. Marginal cost and marginal revenue are derivatives of stated continuous cost and revenue models. They are instantaneous rates, not automatically exact discrete-unit changes; the model is smooth, and the business counts whole units.
A rectangle has width x and length ten minus x, under a fixed perimeter of twenty units, so the feasible domain is x between zero and ten. The area is A of x equals x times, open ten minus x. Differentiating, A prime equals ten minus two x, so the stationary candidate is x equal to five. The second derivative is minus two, negative everywhere, so the candidate is a local maximum; and because the area approaches zero at both domain ends, it is the global maximum. The rectangle is a five by five square with area twenty-five. Every step is there: one variable, stated domain, derivative, candidate, justification. Use practice sheet three point three to test the method, domain and interpretation against its solutions.
Pause and choose. The justification has two parts. Locally, the second derivative at five is minus two, negative, so the point is a maximum. Globally, the area approaches zero at both ends of the domain, so the interior maximum beats the boundary. Stationary points are not always maxima, the middle of a domain proves nothing, and squares are optimal only for this fixed-perimeter area question, not as a law. The marks live in the two-part justification, so write both parts.
Pause for four checks. First, the derivative of x squared is two x, giving tangent slope two at x equal one, while the secant slope from one one to two four is three; average and instantaneous slopes are different questions. Second, the integral of y equals x over minus one to one is zero because the signed contributions cancel, while the geometric area is one. Third, the single most costly omission is the plus c on an indefinite integral. Fourth, the fixed perimeter makes length plus width equal ten, so length is ten minus x and area becomes a function of one variable, ready to differentiate. Carry the order, known, symbols, substitution, justification, into your next practice sheet.