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Implicit and parametric differentiation questions
Plenty of curves have no equation of the form y = f(x), circles first among them, and calculus still applies. Differentiate an equation exactly as it stands, remembering that y depends on x, or divide two parametric rates, and the gradient follows without any rearranging.
8 original questions · 33 marks · the implicit and parametric differentiation notes · Differentiation
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Differentiate y3 with respect to x, and then differentiate the product xy with respect to x.
Worked 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, and these two moves are the whole of implicit differentiation.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).
Worked 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.Find the gradient of the circle x2 + y2 = 100 at the point (6, 8).
Worked 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.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).
Worked 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. Treating xy as a single power and writing its derivative as x dy/dx alone throws away the A mark and every mark after it.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.
Worked 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. One clock drives both coordinates and the gradient is one rate divided by the other, so no Cartesian conversion is needed at any stage.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.
Worked 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. Setting the whole fraction to zero by clearing the denominator instead is the wrong instinct, since it is the numerator alone that controls a zero gradient. The denominator vanishing would give a vertical tangent, which is the question you get asked next.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.
Worked 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. Rationalising the gradient before forming the equation saves most of the algebraic mess at the end.Given that x = sin y, where −π/2 ≤ y ≤ π/2, show that dy/dx = 1/√(1 − x2).
Worked 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. The stated range is not decoration. Without it the square root could take either sign, and this is the derivative of arcsin obtained by running a rate backwards.
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