Exercise 5.6 answers: Continuity and Differentiability

Open NCERT p. 137Matches NCERT’s answer

1. Given: x = 2at², y = at⁴. Find dy/dx.
2. Differentiate x = 2at² with respect to t: dx/dt = 4at.
3. Differentiate y = at⁴ with respect to t: dy/dt = 4at³.
4. Use dy/dx = (dy/dt) ÷ (dx/dt).
5. Divide: dy/dx = 4at³ ÷ 4at = t².

Answerdy/dx = t², for t ≠ 0.

Watch this explained “A parabola, and a constant that cancels”, 8:58 into Curves given through a parameter, and the derivative recovered by the chain rule

Open NCERT p. 137Matches NCERT’s answer

1. Differentiate x = a cos θ with respect to θ: dx/dθ = −a sin θ.
2. Differentiate y = b cos θ with respect to θ: dy/dθ = −b sin θ.
3. dy/dx = dy/dθ ÷ dx/dθ = (−b sin θ) ÷ (−a sin θ) = b/a.

Answerdy/dx = b/a — a constant.

Watch this explained “A pair that is not a curve”, 16:49 into Curves given through a parameter, and the derivative recovered by the chain rule

Open NCERT p. 137Matches NCERT’s answer

1. Differentiate x = sin t: dx/dt = cos t.
2. Differentiate y = cos 2t: dy/dt = −2 sin 2t, which is −4 sin t cos t using sin 2t = 2 sin t cos t.
3. dy/dx = dy/dt ÷ dx/dt = (−4 sin t cos t) ÷ cos t.
4. Cancel cos t (valid where cos t ≠ 0): dy/dx = −4 sin t.

Answerdy/dx = −4 sin t, for cos t ≠ 0.

Watch this explained “The chain rule, divided through”, 2:11 into Curves given through a parameter, and the derivative recovered by the chain rule

Open NCERT p. 137Matches NCERT’s answer

1. Differentiate x = 4t: dx/dt = 4.
2. Differentiate y = 4/t = 4t⁻¹: dy/dt = −4t⁻² = −4/t².
3. dy/dx = dy/dt ÷ dx/dt = (−4/t²) ÷ 4 = −1/t².

Answerdy/dx = −1/t², for t ≠ 0.

Watch this explained “The chain rule, divided through”, 2:11 into Curves given through a parameter, and the derivative recovered by the chain rule

Open NCERT p. 137Matches NCERT’s answer

1. Differentiate x = cos θ − cos 2θ: dx/dθ = −sin θ + 2 sin 2θ.
2. Differentiate y = sin θ − sin 2θ: dy/dθ = cos θ − 2 cos 2θ.
3. dy/dx = dy/dθ ÷ dx/dθ = (cos θ − 2 cos 2θ) ÷ (2 sin 2θ − sin θ).

Answerdy/dx = (cos θ − 2 cos 2θ) ÷ (2 sin 2θ − sin θ).

Watch this explained “The chain rule, divided through”, 2:11 into Curves given through a parameter, and the derivative recovered by the chain rule

Open NCERT p. 137Matches NCERT’s answer

1. Differentiate x = a(θ − sin θ): dx/dθ = a(1 − cos θ).
2. Differentiate y = a(1 + cos θ): dy/dθ = −a sin θ.
3. dy/dx = dy/dθ ÷ dx/dθ = −sin θ ÷ (1 − cos θ).
4. Use the half-angle identities sin θ = 2 sin(θ/2)cos(θ/2) and 1 − cos θ = 2 sin²(θ/2).
5. Simplify: dy/dx = −cos(θ/2) ÷ sin(θ/2) = −cot(θ/2).

Answerdy/dx = −cot(θ/2), for θ not a multiple of 2π.

Watch this explained “The cycloid, and a half angle”, 11:08 into Curves given through a parameter, and the derivative recovered by the chain rule

Open NCERT p. 137Matches NCERT’s answer

1. Given: x = sin³t/√(cos 2t), y = cos³t/√(cos 2t). Find dy/dx.
2. Write x = sin³t·(cos 2t)^−1/2 and y = cos³t·(cos 2t)^−1/2. By the chain rule, d/dt (cos 2t)^−1/2 = (−1/2)(cos 2t)^−3/2·(−2 sin 2t) = sin 2t·(cos 2t)^−3/2.
3. Product rule: dx/dt = 3sin²t·cos t·(cos 2t)^−1/2 + sin³t·sin 2t·(cos 2t)^−3/2 = [3sin²t·cos t·cos 2t + sin³t·sin 2t] ÷ (cos 2t)^3/2.
4. Put sin 2t = 2 sin t cos t in the top: it becomes sin²t·cos t·(3cos 2t + 2sin²t). Since 2sin²t = 1 − cos 2t, the bracket is 1 + 2cos 2t, and sin t·(1 + 2cos 2t) = 3 sin t − 4 sin³t = sin 3t. So dx/dt = sin t·cos t·sin 3t ÷ (cos 2t)^3/2.
5. Product rule again: dy/dt = −3cos²t·sin t·(cos 2t)^−1/2 + cos³t·sin 2t·(cos 2t)^−3/2 = [−3cos²t·sin t·cos 2t + cos³t·sin 2t] ÷ (cos 2t)^3/2.
6. Put sin 2t = 2 sin t cos t in the top: it becomes sin t·cos²t·(2cos²t − 3cos 2t). Since 2cos²t = 1 + cos 2t, the bracket is 1 − 2cos 2t, and cos t·(2cos 2t − 1) = 4cos³t − 3cos t = cos 3t. So dy/dt = −sin t·cos t·cos 3t ÷ (cos 2t)^3/2.
7. Divide: dy/dx = (dy/dt) ÷ (dx/dt) = −sin t·cos t·cos 3t ÷ (sin t·cos t·sin 3t) = −cos 3t/sin 3t, where sin t, cos t and sin 3t are not 0.

Answerdy/dx = −cot 3t

Watch this explained “The chain rule, divided through”, 2:11 into Curves given through a parameter, and the derivative recovered by the chain rule

Open NCERT p. 137Matches NCERT’s answer

1. Given: x = a(cos t + log tan(t/2)), y = a sin t. Find dy/dx.
2. Differentiate y = a sin t: dy/dt = a cos t.
3. For log tan(t/2), use the chain rule: d/dt log tan(t/2) = [1/tan(t/2)]·sec²(t/2)·(1/2) = 1 ÷ [2 sin(t/2)·cos(t/2)] = 1/sin t, because 2 sin(t/2)·cos(t/2) = sin t.
4. So dx/dt = a(−sin t + 1/sin t) = a(1 − sin²t)/sin t = a cos²t/sin t.
5. Divide: dy/dx = (dy/dt) ÷ (dx/dt) = a cos t ÷ (a cos²t/sin t) = sin t/cos t = tan t, where cos t ≠ 0.

Answerdy/dx = tan t

Watch this explained “The chain rule, divided through”, 2:11 into Curves given through a parameter, and the derivative recovered by the chain rule

Open NCERT p. 137Matches NCERT’s answer

1. Differentiate x = a sec θ: dx/dθ = a sec θ tan θ.
2. Differentiate y = b tan θ: dy/dθ = b sec²θ.
3. dy/dx = dy/dθ ÷ dx/dθ = b sec²θ ÷ (a sec θ tan θ) = (b/a)·(sec θ / tan θ).
4. Simplify sec θ / tan θ = 1/sin θ.

Answerdy/dx = (b/a)·(1/sin θ), that is (b/a)·cosec θ.

Watch this explained “The chain rule, divided through”, 2:11 into Curves given through a parameter, and the derivative recovered by the chain rule

Open NCERT p. 137Matches NCERT’s answer

1. Differentiate x = a(cos θ + θ sin θ) using the product rule on θ sin θ: dx/dθ = a(−sin θ + sin θ + θ cos θ) = aθ cos θ.
2. Differentiate y = a(sin θ − θ cos θ) using the product rule on θ cos θ: dy/dθ = a(cos θ − cos θ + θ sin θ) = aθ sin θ.
3. dy/dx = dy/dθ ÷ dx/dθ = (aθ sin θ) ÷ (aθ cos θ).
4. Cancel aθ: dy/dx = sin θ / cos θ = tan θ.

Answerdy/dx = tan θ, for θ ≠ 0.

Watch this explained “The chain rule, divided through”, 2:11 into Curves given through a parameter, and the derivative recovered by the chain rule

Open NCERT p. 137One way to think about it

1. To show: dy/dx = −y/x.
2. Here a > 0, a ≠ 1 and −1 < t < 1. Write x = a^{(1/2)·sin⁻¹t} and y = a^{(1/2)·cos⁻¹t}, since a square root is the power 1/2.
3. Take log of x: log x = (1/2)(sin⁻¹t)(log a). Differentiate with respect to t: (1/x)(dx/dt) = (log a) ÷ (2√(1−t²)).
4. Take log of y: log y = (1/2)(cos⁻¹t)(log a). Differentiate, using d/dt cos⁻¹t = −1/√(1−t²): (1/y)(dy/dt) = −(log a) ÷ (2√(1−t²)).
5. So dx/dt = x·(log a)/(2√(1−t²)) and dy/dt = −y·(log a)/(2√(1−t²)).
6. Divide: dy/dx = (dy/dt) ÷ (dx/dt). The common factor (log a)/(2√(1−t²)) is not 0, so it cancels, leaving dy/dx = −y/x.

In shortdy/dx = −y/x, as required.

Watch this explained “An answer in the coordinates”, 19:20 into Curves given through a parameter, and the derivative recovered by the chain rule