Questions on implicit and parametric differentiation

Question 1

Differentiate y³ with respect to x, and then differentiate the product xy with respect to x.

Accepted Answer

y³ gives 3y² dy/dx. Differentiate as though y were the variable, then attach dy/dx by the chain rule. The product xy needs the product rule, giving y + x dy/dx. B1 B1, one for each result.

Question 2

A curve has equation y² = 4x. Find dy/dx in terms of y, and hence find the gradient of the curve at the point (4, 4).

Accepted Answer

Differentiating both sides with respect to x gives 2y dy/dx = 4, so dy/dx = 2/y. At (4, 4) the gradient is 2/4 = ½. M1 for the implicit step, A1 for the gradient. Notice the answer depends only on y, so the curve has two different gradients above and below the x-axis at the same value of x.

Question 3

A curve has equation x = y³ + 2y. Find dx/dy, and hence write down dy/dx in terms of y.

Accepted Answer

Differentiating with respect to y, dx/dy = 3y² + 2. B1. Turning the rate upside down, dy/dx = 1/(dx/dy) = 1/(3y² + 2). B1. The reciprocal step is valid where dx/dy ≠ 0, and 3y² + 2 is at least 2 for every real y, so this curve has a gradient at every point and never a vertical tangent. The equation gives x in terms of y, and rearranging it into the form y = f(x) would mean solving a cubic; differentiating with respect to y and inverting reaches the gradient without that.

Question 4

A curve has parametric equations x = t³, y = 4t. Find dy/dx in terms of t, and state the value of t at which the curve has a vertical tangent, giving a reason.

Accepted Answer

dx/dt = 3t² and dy/dt = 4, so dy/dx = (dy/dt)/(dx/dt) = 4/(3t²), valid wherever t ≠ 0. M1 for both rates, A1 for the quotient. At t = 0 the rate dx/dt is zero while dy/dt = 4 is not, so at that instant the point is moving vertically and the tangent is vertical. B1 for t = 0 with the reason.

Question 5

Find the gradient of the circle x² + y² = 100 at the point (6, 8).

Accepted Answer

Differentiate every term where it stands: 2x + 2y dy/dx = 0, so dy/dx = −x/y and at (6, 8) the gradient is −6/8 = −3/4. M1 for the implicit differentiation, A1 for 2x + 2y dy/dx = 0, M1 for the rearrangement, A1 for −3/4. As a check, the radius to (6, 8) has gradient 8/6 = 4/3, and the two gradients multiply to −1, so the calculus agrees with the perpendicular-radius geometry.

Question 6

The curve C has equation 2x² + 3y² − xy = 16. Find dy/dx in terms of x and y, and hence find the gradient of C at the point (2, 2).

Accepted Answer

Differentiating term by term gives 4x + 6y dy/dx − (y + x dy/dx) = 0, where the bracket comes from the product rule on xy. Gathering the dy/dx terms: dy/dx (6y − x) = y − 4x, so dy/dx = (y − 4x)/(6y − x). At (2, 2), which does lie on C since 8 + 12 − 4 = 16, the gradient is (2 − 8)/(12 − 2) = −6/10 = −3/5. M1 for the implicit differentiation, A1 for the product-rule term, M1 for gathering, A1 for the expression, A1 for −3/5. The term xy is a product, so its derivative is y + x dy/dx; x dy/dx alone omits the first term.

Question 7

A curve has parametric equations x = t², y = t³. Find dy/dx in terms of t, and find the coordinates and the gradient of the point on the curve where t = 2.

Accepted Answer

dy/dx = (dy/dt)/(dx/dt) = 3t²/(2t) = 3t/2, valid for t ≠ 0. At t = 2 the point is (4, 8) and the gradient is 3. M1 for both derivatives, A1 for 3t/2, B1 for the point, A1 for the gradient.

Question 8

The curve C has equation x² + 4y² = 40. Show that the point P(2, 3) lies on C, and find an equation of the normal to C at P.

Accepted Answer

At P the left side is 4 + 4 × 9 = 40, which is the right side, so P lies on C. B1. Differentiating every term with respect to x, 2x + 8y dy/dx = 0, so dy/dx = −x/(4y) for y ≠ 0. M1 for the implicit differentiation, A1 for the expression. At P this is −2/12 = −1/6, so the tangent has gradient −1/6 and the normal, being perpendicular, has gradient 6. M1 for the negative reciprocal used with point-gradient form. The normal is y − 3 = 6(x − 2), that is y = 6x − 9. A1. As a check, 6 × 2 − 9 = 3, so the line does pass through P.

Question 9

A curve has parametric equations x = t − sin t, y = 1 − cos t, for 0 < t < 2π. Find dy/dx in terms of t, and find the exact gradient of the curve at the point where t = π/3.

Accepted Answer

dx/dt = 1 − cos t and dy/dt = sin t, so dy/dx = (sin t)/(1 − cos t). M1 for both rates, A1 for the quotient. Strictly inside 0 < t < 2π the value of cos t is less than 1, so 1 − cos t is never zero there and the division is valid throughout. At t = π/3, sin (π/3) = √3/2 and cos (π/3) = ½, so the denominator is 1 − ½ = ½. M1 for both exact values. The gradient is (√3/2) ÷ ½ = √3. A1. The stated interval is what keeps the denominator away from zero, since cos t = 1 at both t = 0 and t = 2π, where this curve has a vertical tangent.

Question 10

A curve has parametric equations x = 2t, y = 4/t, for t ≠ 0. Find dy/dx in terms of t, show that the Cartesian equation of the curve is xy = 8, and verify by differentiating xy = 8 implicitly that the two routes give the same gradient.

Accepted Answer

dx/dt = 2 and dy/dt = −4/t², so dy/dx = (−4/t²)/2 = −2/t². M1 for both rates, A1 for the quotient. Multiplying the two parametric equations removes t in one step: xy = 2t × (4/t) = 8, so the Cartesian equation is xy = 8. M1 for eliminating the parameter, A1. Differentiating xy = 8 with respect to x, the product rule gives y + x dy/dx = 0, so dy/dx = −y/x. In terms of t that is −(4/t)/(2t) = −4/(2t²) = −2/t², the same expression as before. A1 for the implicit derivative with the comparison.

Question 11

A curve has parametric equations x = t² + 1, y = t³ − 3t. Find the coordinates of the two points on the curve at which the tangent is parallel to the x-axis.

Accepted Answer

A tangent parallel to the x-axis has dy/dx = 0, which on a parametric curve means dy/dt = 0 while dx/dt is not zero. M1 for setting dy/dt to zero. Here dy/dt = 3t² − 3 = 3(t − 1)(t + 1), which vanishes at t = 1 and at t = −1. A1 for both values. At those values dx/dt = 2t takes the values 2 and −2, neither of them zero, so the gradient there is genuinely zero rather than undefined. M1 for checking dx/dt. Substituting, t = 1 gives x = 2 and y = 1 − 3 = −2, and t = −1 gives x = 2 and y = −1 + 3 = 2, so the points are (2, −2) and (2, 2). A1. Both points stand at x = 2, one above the other, because the curve doubles back on itself; the two values of t are what separate them.

Question 12

The curve C has equation x² + xy + y² = 12. Find the coordinates of the two points on C at which the tangent is parallel to the x-axis.

Accepted Answer

Differentiating gives 2x + (y + x dy/dx) + 2y dy/dx = 0, so dy/dx (x + 2y) = −(2x + y) and dy/dx = −(2x + y)/(x + 2y). A tangent parallel to the x-axis needs dy/dx = 0, so the numerator must vanish: 2x + y = 0, that is y = −2x. Substituting into the equation of C gives x² − 2x² + 4x² = 12, so 3x² = 12 and x = ±2. The points are (2, −4) and (−2, 4). M1 implicit differentiation, A1 correct, M1 rearrange for dy/dx, A1 expression, M1 set the numerator to zero and substitute, A1 both points. The denominator vanishing would give a vertical tangent instead.

Question 13

A curve has parametric equations x = 3 cos t, y = 2 sin t, 0 ≤ t < 2π. Find an equation of the tangent to the curve at the point where t = π/3, giving your answer in the form ax + by = c with a, b and c exact.

Accepted Answer

dx/dt = −3 sin t and dy/dt = 2 cos t, so dy/dx = −(2 cos t)/(3 sin t). At t = π/3 this is −(2 × ½)/(3 × √3/2) = −2/(3√3) = −2√3/9. The point is (3 cos π/3, 2 sin π/3) = (3/2, √3). The tangent is y − √3 = −(2/(3√3))(x − 3/2). Multiplying through by 3√3 gives 3√3 y − 9 = −2x + 3, so 2x + 3√3 y = 12. M1 for both derivatives, M1 for the ratio, A1 for the gradient, B1 for the point, M1 for the straight-line equation, A1 for the final form. Check by substituting the point back: 3 + 3√3 × √3 = 3 + 9 = 12.

Question 14

Given that x = sin y, where −π/2 ≤ y ≤ π/2, show that dy/dx = 1/√(1 − x²).

Accepted Answer

Differentiate with respect to y rather than x, giving dx/dy = cos y. Turning the rate upside down, dy/dx = 1/cos y. Now cos y = √(1 − sin² y) = √(1 − x²), where the positive root is correct because cos y ≥ 0 on the stated range. Hence dy/dx = 1/√(1 − x²). M1 for dx/dy, M1 for the reciprocal, M1 for the Pythagorean replacement, A1 for the conclusion. This is the derivative of arcsin x.

Question 15

The curve C has equation √x + √y = 3, for 0 < x < 9. Show that dy/dx = −√y/√x. The tangent to C at the point (a, b) on C meets the x-axis at one point and the y-axis at another. Show that the two intercepts add to 9, whichever point of C the tangent touches.

Accepted Answer

Write the equation as x^1/2 + y^1/2 = 3 and differentiate every term with respect to x. The second term is a function of y, so the chain rule attaches dy/dx: ½x^−1/2 + ½y^−1/2 dy/dx = 0. M1 for differentiating both roots with dy/dx on the second. Multiplying by 2 and rearranging, dy/dx = −y^1/2/x^1/2 = −√y/√x. A1. At (a, b) the gradient is −√b/√a, so the tangent is y − b = −(√b/√a)(x − a). M1 for the tangent at a general point of C. Putting y = 0 gives −b = −(√b/√a)(x − a), so x − a = b√a/√b = √a√b and the x-intercept is a + √(ab). A1. Putting x = 0 gives y − b = −(√b/√a)(0 − a) = a√b/√a = √a√b, so the y-intercept is b + √(ab). A1. Adding them, a + b + 2√(ab) = (√a + √b)², and (a, b) lies on C, so √a + √b = 3 and the sum is 9 for every such point. A1. The restriction 0 < x < 9 keeps both a and b positive, so the tangent is neither horizontal nor vertical and both intercepts exist.