Application of Derivatives — Practice

Question 1

A ladder 5 m long leans against a vertical wall. Its foot slides away from the wall at 2 m per second. How fast is the top sliding down when the foot is 3 m from the wall?

Accepted Answer

Let x be the distance of the foot from the wall and y the height of the top. Pythagoras gives x squared plus y squared equal to 25. The length is genuinely constant so 25 may stay, but x and y must remain symbols. Differentiating with respect to time gives 2x times dx by dt plus 2y times dy by dt equal to zero. Now substitute: when x is 3 the relation gives y equal to 4, and dx by dt is 2. So 2 times 3 times 2, which is 12, plus 8 times dy by dt equals zero, giving dy by dt equal to minus 1.5. The top is sliding down at 1.5 m per second. The negative sign is the answer reporting that the height decreases, so it should not be repeated as a direction word.

Question 2

Air is pumped into a spherical balloon so that its volume increases at 100 cubic cm per second. Find the rate of increase of the radius when the radius is 10 cm.

Accepted Answer

The volume of a sphere is four thirds pi r cubed. Differentiate with respect to time, keeping r as a symbol: dV by dt equals 4 pi r squared times dr by dt, by the chain rule. Only now substitute the instantaneous values, dV by dt equal to 100 and r equal to 10. That gives 100 equal to 4 pi times 100 times dr by dt, so dr by dt equals 100 divided by 400 pi, which is 1 over 4 pi centimetres per second, approximately 0.08. Substituting r equal to 10 before differentiating would have made the volume a constant and produced the nonsensical conclusion that the rate is zero.

Question 3

The side of a cube is measured as 10 cm with a possible error of 0.02 cm. Find the approximate error in the calculated volume, and the percentage error.

Accepted Answer

The volume is the side cubed, so the derivative with respect to the side is three times the side squared. The approximate error in volume is that derivative times the error in the side, which is 3 times 100 times 0.02, giving 6 cubic cm. For the percentage figures, the side is uncertain by 0.02 out of 10, which is 0.2 per cent, while the volume is uncertain by 6 out of 1000, which is 0.6 per cent. The volume error is three times the side error, and the factor of three is precisely the exponent. The general pattern is that a percentage error is multiplied by the power to which the quantity is raised.

Question 4

Find the intervals on which f(x) equal to 2x cubed minus 9x squared plus 12x plus 5 is increasing and decreasing.

Accepted Answer

Differentiating gives 6x squared minus 18x plus 12. Factorising completely, which is what makes the sign analysis possible, gives 6 times (x minus 1) times (x minus 2). The derivative is zero at x equal to 1 and x equal to 2, dividing the line into three intervals. Testing one point in each: at x equal to 0 the value is 6 times minus 1 times minus 2, which is 12 and positive; at x equal to 1.5 it is 6 times 0.5 times minus 0.5, which is negative; at x equal to 3 it is 6 times 2 times 1, which is positive. So the function increases on the interval below 1 and on the interval above 2, and decreases between 1 and 2. The two increasing intervals must be reported separately, since joining them would claim the function rises from 0 to 3.

Question 5

Show that f(x) equal to x cubed is strictly increasing on the whole real line even though its derivative vanishes at the origin.

Accepted Answer

The derivative is 3x squared, which is positive for every non-zero x and zero only at the origin. Strict increase requires that whenever one input exceeds another the corresponding output does too, and a derivative that is positive everywhere except at isolated points guarantees this. The graph never levels off over any interval; it merely has a single instant of zero slope at the origin before continuing to rise. Directly, if a is less than b then b cubed minus a cubed factors as (b minus a) times (b squared plus ab plus a squared), and the second factor is positive for all real a and b not both zero, so the difference is positive. This is why a single stationary point does not prevent strict monotonicity.

Question 6

Find the values of k for which f(x) equal to x cubed minus 3kx squared plus 9x is increasing for all real x.

Accepted Answer

Differentiating gives 3x squared minus 6kx plus 9. For the function to increase everywhere this quadratic must be non-negative for every real x. A quadratic with a positive leading coefficient is non-negative everywhere exactly when its discriminant is at most zero. The discriminant is 36k squared minus 4 times 3 times 9, which is 36k squared minus 108. Requiring this to be at most zero gives k squared at most 3, so k lies between minus root 3 and root 3 inclusive. At the endpoints the derivative touches zero at a single point, which as with x cubed does not prevent strict increase, so the closed interval is correct.

Question 7

Find the local maximum and minimum values of 2x cubed minus 9x squared plus 12x plus 5, and comment on their relative sizes.

Accepted Answer

The derivative factorises as 6 times (x minus 1) times (x minus 2), so the critical points are x equal to 1 and x equal to 2. By the first derivative test the sign goes positive to negative at 1, giving a local maximum, and negative to positive at 2, giving a local minimum. Evaluating, f(1) is 2 minus 9 plus 12 plus 5, which is 10, and f(2) is 16 minus 36 plus 24 plus 5, which is 9. The second derivative is 12x minus 18, which is minus 6 at x equal to 1 confirming a maximum, and plus 6 at x equal to 2 confirming a minimum. Note that the local maximum value 10 exceeds the local minimum value 9, which is normal: local means compared with neighbours, not compared with everything.

Question 8

Examine x to the fourth and x cubed at the origin, and explain why the second derivative test cannot distinguish them.

Accepted Answer

For x to the fourth, the first derivative is 4x cubed, which vanishes at the origin, so it is a critical point, and the second derivative is 12x squared, which also vanishes there. The test is inconclusive, so use the first derivative test: 4x cubed is negative for negative x and positive for positive x, a change from negative to positive, so the origin is a local minimum. For x cubed, the first derivative is 3x squared, which vanishes at the origin, and the second derivative is 6x, which also vanishes there. Again inconclusive, but now 3x squared is positive on both sides, so there is no sign change and the origin is not an extremum at all; it is a point of inflection. Both functions give the identical reading of zero yet have different answers, which is exactly why a zero second derivative can carry no conclusion.

Question 9

Find the absolute maximum and minimum values of f(x) equal to x cubed minus 3x on the closed interval from minus 2 to 3.

Accepted Answer

The derivative is 3x squared minus 3, which factorises as 3 times (x minus 1) times (x plus 1), so the critical points are x equal to minus 1 and x equal to 1, both inside the interval. Now evaluate at the critical points and at both endpoints. At minus 2 the value is minus 8 plus 6, which is minus 2. At minus 1 it is minus 1 plus 3, which is 2. At 1 it is 1 minus 3, which is minus 2. At 3 it is 27 minus 9, which is 18. Comparing the four values, the absolute maximum is 18 at the endpoint x equal to 3, and the absolute minimum is minus 2, attained at both x equal to minus 2 and x equal to 1. Two points are worth noting: the maximum sits at an endpoint where the derivative is not zero, and the minimum is attained twice, which is a valid answer.

Question 10

Squares of side x are cut from the corners of a 24 cm square sheet and the flaps folded up to form an open box. Find x for maximum volume, and the maximum volume.

Accepted Answer

Cutting x from both ends of each side leaves a base of side 24 minus 2x, and the height is x, so the volume is x times the square of (24 minus 2x). The expression is already in one variable, so state the domain: we need x positive and 24 minus 2x positive, giving x strictly between 0 and 12. Differentiating by the product rule gives the square of (24 minus 2x) plus x times 2 times (24 minus 2x) times minus 2. Factoring out the common (24 minus 2x) gives (24 minus 2x) times the quantity (24 minus 2x minus 4x), which is (24 minus 2x)(24 minus 6x). Setting this to zero gives x equal to 12 or x equal to 4, and the first is excluded by the domain since it leaves no base. So x equals 4, and the volume is 4 times 16 squared, which is 1024 cubic cm. Just below 4 the derivative is positive and just above it is negative, confirming a maximum.

Question 11

A rectangle has perimeter 40 cm. Find the dimensions giving the greatest area, using both calculus and the AM-GM inequality.

Accepted Answer

Let the sides be x and y, so the constraint is 2x plus 2y equal to 40, that is x plus y equal to 20. The area is xy, and using the constraint to eliminate y gives area equal to x times (20 minus x), with domain x strictly between 0 and 20. Differentiating gives 20 minus 2x, which vanishes at x equal to 10, and the second derivative is minus 2, negative, confirming a maximum. So both sides are 10 cm and the area is 100 square cm. By AM-GM the same follows in one line: since x plus y is fixed at 20, the geometric mean of x and y is at most their arithmetic mean of 10, so the product xy is at most 100, with equality exactly when x equals y. The inequality route needs no calculus at all.

Question 12

A curve has second derivative equal to 6x minus 12. Find where it is concave up, where concave down, and locate any point of inflection.

Accepted Answer

The second derivative describes concavity: positive means concave up and negative means concave down. Here 6x minus 12 is positive when x exceeds 2 and negative when x is less than 2, so the curve is concave down to the left of x equal to 2 and concave up to the right of it. A point of inflection requires the second derivative to change sign, not merely to vanish. Here it vanishes at x equal to 2 and genuinely changes from negative to positive there, so x equal to 2 is a point of inflection. Had the second derivative been something like 6 times the square of (x minus 2), it would vanish at x equal to 2 without changing sign, and there would be no inflection despite the zero.

Practice — Application of Derivatives

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