Free SACE Mathematical Methods Flashcard Review

For a=e the limit is 1. This makes the derivative of e^x equal to e^x. Numerical estimates should approach the limit from small positive and negative h rather than substitute h=0 into the undefined quotient.

Use the chain rule: y′=−14e^(−2x). The factor −2 comes from differentiating the exponent. Since the exponential is positive, this function is decreasing everywhere.

The ratio is 0.03 per unit time, a constant relative growth rate. The absolute growth rate is 3.6e^(0.03t), which increases with the population. A continuous relative rate is not identical to the finite percentage increase over a whole time unit.

At x=0 the point is (0,1) and the derivative is 1. Hence y−1=x, or y=x+1. The tangent agrees with the curve locally but is not the same function.

The derivative remains e^(x−4) because the derivative of x−4 is 1. The graph is shifted four units right relative to e^x. A shift is not a reason to introduce an extra factor of 4.

The outer exponential differentiates to itself, but the inner function has derivative 3. Therefore the result is 3e^(3x). Omitting this factor confuses the derivative with respect to the exponent and the derivative with respect to x.

A(0)=80 and A′(0)=−16 amount-units per time-unit. The negative rate describes decay, not a negative amount. The amount stays positive for all finite real t.

No: Ae^(kt)>0 for A>0 and every finite real t. With k<0 it approaches zero as t increases without bound. A practical model can cease to be appropriate before that mathematical limit is relevant.

Positive k gives increasing y, negative k gives decreasing y, and k=0 gives a constant. The derivative’s magnitude is proportional to y. This describes continuous proportional change, not addition of a fixed amount each period.

The input must be in radians. If an angle variable is measured in degrees, conversion introduces a factorπ/180. State or check the convention before differentiating a trigonometric model.

The derivative is 12cos(3x−1). Differentiate the outer sine and multiply by the inner derivative 3. The amplitude 4 remains a constant multiplier.

The derivative is −10sin(2t). The vertical shift 7 changes the level but not the rate of change. The minus sign comes from the cosine derivative, while the factor 2 comes from the inner function.

The point is (π/2,1) and the slope cos(π/2)=0. The tangent is the horizontal line y=1. A zero derivative indicates a stationary point, whose type requires further reasoning.

The period is 2π/(π/4)=8 time-units. The amplitude is 2 and the midline is 6. The derivative’s coefficient should not be used as the period.

h′(t)=(π/2)cos(πt/4), so its greatest possible value isπ/2. This occurs when the cosine equals 1, provided such a time is in the model domain. The greatest rate is not the greatest height.

Velocity is −6sin(2t) and acceleration is −12cos(2t). Differentiate with respect to time twice, retaining the chain-rule factor each time. Their units are position-units per time-unit and per time-unit squared respectively.

Differentiate, solve for zeros of the derivative in that domain and test its sign between them. Positive derivative means increasing output. Include all periodic solutions in the domain rather than stopping at a calculator’s principal solution.

At a smooth turning point the tangent is horizontal, so the first derivative is zero. The model changes from rising to falling or the reverse. The derivative changes sign; merely finding a zero derivative is not sufficient for every possible function.

The derivative is cos x−sin x, which equals 1 at x=0. Differentiate the sum term by term. The cosine term contributes zero slope at 0 even though its function value there is 1.

For y=uv, y′=u′v+uv′. Each factor contributes a change while the other remains in the corresponding term. In general u′v′ omits both required contributions and gives a different expression.