Example Set 1 – Tangents and Normals

Practice Questions with Step-by-Step Solution

Question 1

Curve: [y = x^{2}] at [x = -2]. Find tangent slope, normal slope, tangent & normal equations, and [\theta].

Solution:

1. [\dfrac{dy}{dx} = 2x].
2. [m_{t} = 2(-2) = -4].
3. [m_{n} ][= -\dfrac{1}{-4} ][= \dfrac{1}{4}].
4. Point: [y(-2)=4] → [(-2,4)].
5. Tangent eqn: [y – 4 = -4(x + 2)] → [y = -4x -4].
6. Normal eqn: [y – 4 ][= \dfrac{1}{4}(x + 2)] → [y ][= \dfrac{1}{4}x + \dfrac{9}{2}].
7. Angle: [\theta = \tan^{-1}(-4)].

Question 2

Curve: [y = e^{2x}] at [x = 0].

Solution:

1. [\dfrac{dy}{dx} = 2e^{2x}].
2. [m_{t} = 2e^{0} = 2].
3. [m_{n} = -\dfrac{1}{2}].
4. Point: [y(0)=1] → [(0,1)].
5. Tangent eqn: [y – 1 = 2(x – 0)] → [y = 2x + 1].
6. Normal eqn: [y – 1 ][= -\dfrac{1}{2}(x – 0)] → [y ][= -\dfrac{1}{2}x + 1].
7. Angle: [\theta = \tan^{-1}(2)].

Question 3

Curve: [y = \sin x] at [x = \dfrac{\pi}{6}].

Solution:

1. [\dfrac{dy}{dx} = \cos x].
2. [m_{t} ][= \cos(\dfrac{\pi}{6}) ][= \dfrac{\sqrt{3}}{2}].
3. [m_{n} ][= -\dfrac{1}{\sqrt{3}/2} ][= -\dfrac{2}{\sqrt{3}} ][= -\dfrac{2\sqrt{3}}{3}].
4. Point: [y(\dfrac{\pi}{6})][=\dfrac{1}{2}] → [(\dfrac{\pi}{6} \dfrac{1}{2})].
5. Tangent eqn: [y – \dfrac{1}{2} ][= \dfrac{\sqrt{3}}{2}\big(x – \dfrac{\pi}{6}\big)].
6. Normal eqn: [y – \dfrac{1}{2} ][= -\dfrac{2\sqrt{3}}{3}\big(x – \dfrac{\pi}{6}\big)].
7. Angle: [\theta ][= \tan^{-1}\big(\dfrac{\sqrt{3}}{2}\big)].

Question 4

Curve: [y = \ln(x^{2}+1)] at [x=1].

Solution:

1. [\dfrac{dy}{dx} ][= \dfrac{1}{x^{2}+1}\cdot 2x ][= \dfrac{2x}{x^{2}+1}].
2. At [x=1]: [m_{t} = \dfrac{2}{2} = 1].
3. [m_{n} = -1].
4. Point: [y(1)=\ln 2] → [(1,\ln 2)].
5. Tangent eqn: [y – \ln 2 = 1\cdot (x – 1)] → [y = x + \ln 2 -1].
6. Normal eqn: [y – \ln 2 ][= -1(x – 1)] → [y = -x + 1 + \ln 2].
7. Angle: [\theta ][= \tan^{-1}(1) ][= 45^\circ].

Question 5

Implicit curve: [xy + y^{2} = 6] at point [(2,1)].

Solution:

1. Differentiate implicitly w.r.t [x]: [x\dfrac{dy}{dx} + y + 2y\dfrac{dy}{dx} ][= 0].
2. Collect [\dfrac{dy}{dx}]: [\dfrac{dy}{dx}(x + 2y) = -y].
3. [\dfrac{dy}{dx} ][= -\dfrac{y}{x + 2y}].
4. At [(2,1)]: [m_{t} ][= -\dfrac{1}{2 + 2} ][= -\dfrac{1}{4}].
5. [m_{n} ][= -\dfrac{1}{m_{t}} ][= -\dfrac{1}{-1/4} = 4].
6. Tangent eqn: [y – 1 ][= -\dfrac{1}{4}(x – 2)] → [y ][= -\dfrac{1}{4}x + \dfrac{3}{2}].
7. Normal eqn: [y – 1 = 4(x – 2)] → [y = 4x -7].
8. Angle: [\theta ][= \tan^{-1}(-\dfrac{1}{4})].

Question 6

Circle: [x^{2}+y^{2}=50] at point [(-5,5)].

Solution:

1. Differentiate implicitly: [2x + 2y,\dfrac{dy}{dx} = 0] → [\dfrac{dy}{dx} = -\dfrac{x}{y}].
2. At (-5,5): [m_{t} = -\dfrac{-5}{5} = 1].
3. [m_{n} = -1].
4. Tangent eqn: [y – 5 = 1(x +5)] → [y = x +10].
5. Normal eqn: [y – 5 = -1(x +5)] → [y = -x].
6. Angle: [\theta = \tan^{-1}(1) = 45^\circ].

Question 7

Parametric curve: [x = t^{2}], [y = t^{3}] at [t=2]. Find slope of tangent and normal and equations in [x,y]-form.

Solution:

1. Compute parametric derivatives: [\dfrac{dx}{dt} = 2t], [\dfrac{dy}{dt} = 3t^{2}].
2. Slope of tangent: [\dfrac{dy}{dx} ][= \dfrac{dy/dt}{dx/dt} ][= \dfrac{3t^{2}}{2t} ][= \dfrac{3t}{2}].
3. At [t=2]: [m_{t} ][= \dfrac{3\cdot 2}{2} = 3].
4. [m_{n} = -\dfrac{1}{3}].
5. Point: [x(2)=4], [y(2)=8] → [(4,8)].
6. Tangent eqn: [y – 8 = 3(x – 4)] → [y = 3x -4].
7. Normal eqn: [y – 8 = -\dfrac{1}{3}(x – 4)] → [y = -\dfrac{1}{3}x + \dfrac{28}{3}].
8. Angle: [\theta = \tan^{-1}(3)].

Question 8

Curve: [y = \sqrt{x}] at [x = 4].

Solution:

1. [\dfrac{dy}{dx} ][= \dfrac{1}{2\sqrt{x}}].
2. At [x=4]: [m_{t} ][= \dfrac{1}{2\cdot 2} ][= \dfrac{1}{4}].
3. [m_{n} = -4].
4. Point: [y(4) = 2] → [(4,2)].
5. Tangent eqn: [y – 2 ][= \dfrac{1}{4}(x – 4)] → [y ][= \dfrac{1}{4}x + 1].
6. Normal eqn: [y – 2 = -4(x – 4)] → [y = -4x + 18].
7. Angle: [\theta = \tan^{-1}(\dfrac{1}{4})].

Question 9

Curve: [y = \tan x] at [x = \dfrac{\pi}{8}].

Solution:

1. [\dfrac{dy}{dx} = \sec^{2}x].
2. At [x=\dfrac{\pi}{8}]: [m_{t} ][= \sec^{2}(\dfrac{\pi}{8})]. (Keep exact as [\sec^{2}(\dfrac{\pi}{8})].)
3. Slope of normal: [m_{n} ][= -\dfrac{1}{\sec^{2}(\dfrac{\pi}{8})} ][= -\cos^{2}(\dfrac{\pi}{8})].
4. Point: [y(\dfrac{\pi}{8}) ][= \tan(\dfrac{\pi}{8})] → [(\dfrac{\pi}{8} \ \tan(\dfrac{\pi}{8}))].
5. Tangent eqn: [y – \tan(\dfrac{\pi}{8}) ][= \sec^{2}(\dfrac{\pi}{8})\big(x – \dfrac{\pi}{8}\big)].
6. Normal eqn: [y – \tan(\dfrac{\pi}{8}) ][= -\cos^{2}(\dfrac{\pi}{8})\big(x – \dfrac{\pi}{8}\big)].
7. Angle: [\theta ][= \tan^{-1}(\sec^{2}(\dfrac{\pi}{8}))].

Question 10

Physics application: position [s(t)=5t^{2}-t] (m) — treat curve [y=s(t)] vs [x=t]. At [t=3] find slope of tangent (velocity), slope of normal and tangent/normal equations in [t,s] coordinates.

Solution:

1. Interpret [x=t], [y=s(t)]. Derivative w.r.t [t] gives slope of tangent in the t–s plane: [\dfrac{dy}{dx} = \dfrac{ds}{dt}].
2. Compute velocity (tangent slope): [\dfrac{ds}{dt} = 10t -1]. At [t=3]: [m_{t} = 10(3)-1 = 29\ \text{m/s}].
3. Slope of normal: [m_{n} = -\dfrac{1}{29}].
4. Point: [s(3)=5(9)-3=45-3=42] → [(t,s)=(3,42)].
5. Tangent eqn (in t–s plane): [s – 42 = 29(t – 3)].
6. Normal eqn: [s – 42 ][= -\dfrac{1}{29}(t – 3)].
7. Angle of motion: [\theta = \tan^{-1}(29)] (steep slope → nearly vertical).