Example Set 1 – Equation of the Tangent and the Normals

Practice Questions with Step-by-Step Solutions

Question 1

Find the equation of the tangent and the normal to [y = x^{2} – 4x + 3] at [x = 2].

Step-by-step solution:

1. Compute derivative: [\dfrac{d}{dx}(y)=2x-4].
2. Slope of tangent at [x=2]: [m_{t}=2(2)-4=0].
3. Point on curve: [y(2)=2^{2}-8+3= -1] → point [(2,-1)].
4. Tangent equation (point-slope): [y+1 = 0\cdot (x-2)] ⇒ [y = -1].
5. Slope of normal: tangent slope 0 ⇒ normal is vertical ⇒ equation [x = 2].

Conclusion: Tangent: [y=-1]. Normal: [x=2].

Question 2

Find tangent and normal to [y=\sin x] at [x=\pi/4].

Step-by-step solution:

1. [\dfrac{d}{dx}(y)=\cos x].
2. [m_{t}][=\cos(\pi/4)][=\dfrac{\sqrt{2}}{2}].
3. Point: [y(\pi/4)][=\dfrac{\sqrt{2}}{2}] → [(\dfrac{\pi}{4},\dfrac{\sqrt{2}}{2})].
4. Tangent: [y – \dfrac{\sqrt{2}}{2} ][= \dfrac{\sqrt{2}}{2}\big(x – \dfrac{\pi}{4}\big)].
5. Normal slope: [m_{n}=-\dfrac{1}{m_{t}} ][= -\dfrac{1}{\sqrt{2}/2} = -\sqrt{2}].
6. Normal: [y – \dfrac{\sqrt{2}}{2} ][= -\sqrt{2}\big(x – \dfrac{\pi}{4}\big)].

Conclusion: Tangent and normal as above.

Question 3

For [y = \ln x], find tangent and normal at [x=1].

Step-by-step solution:

1. [\dfrac{d}{dx}(y)=\dfrac{1}{x}].
2. [m_{t}=\dfrac{1}{1}=1].
3. Point: [y(1)=0] → [(1,0)].
4. Tangent: [y – 0 = 1\cdot(x – 1)] ⇒ [y = x – 1].
5. Normal slope: [m_{n}=-1].
6. Normal: [y = – (x – 1)] ⇒ [y = -x + 1].

Conclusion: Tangent [y = x-1], Normal [y = -x+1].

Question 4

Find tangent and normal to the circle [x^{2}+y^{2}=25] at point [(4,3)].

Step-by-step solution:

1. Implicitly differentiate: [2x + 2y,\dfrac{dy}{dx} = 0] ⇒ [\dfrac{dy}{dx} = -\dfrac{x}{y}].
2. At (4,3): [m_{t} = -\dfrac{4}{3}].
3. Tangent: [y – 3 = -\dfrac{4}{3}(x – 4)].
4. Normal slope: [m_{n} ][= -\dfrac{1}{m_{t}} ][= -\dfrac{1}{-4/3} ][= \dfrac{3}{4}].
5. Normal: [y – 3 = \dfrac{3}{4}(x – 4)].

Conclusion: Tangent and normal as above.

Question 5

For parametric curve [x = t^{2},\ y = t^{3}], find tangent and normal at [t=1] (give equations in x,y).

Step-by-step solution:

1. Compute derivatives: [\dfrac{dx}{dt}=2t], [\dfrac{dy}{dt}=3t^{2}].
2. Slope [\dfrac{dy}{dx}][=\dfrac{3t^{2}}{2t}][=\dfrac{3t}{2}]. At [t=1]: [m_{t}= \dfrac{3}{2}].
3. Point: [x(1)=1,\ y(1)=1] → [(1,1)].
4. Tangent: [y – 1 ][= \dfrac{3}{2}(x – 1)].
5. Normal slope: [m_{n} ][= -\dfrac{2}{3}].
6. Normal: [y – 1 ][= -\dfrac{2}{3}(x – 1)].

Conclusion: Tangent [y -1 ][= \dfrac{3}{2}(x-1)], Normal [y -1 = -\dfrac{2}{3}(x-1)].

Question 6

Find the equation of the tangent and normal to [y = x^{3} – 3x + 1] at [x=1].

Step-by-step solution:

1. [\dfrac{d}{dx}(y)=3x^{2}-3].
2. At [x=1]: [m_{t}=3-3=0].
3. Point: [y(1)=1 -3 +1 = -1] → [(1,-1)].
4. Tangent: horizontal ⇒ [y = -1].
5. Normal: vertical ⇒ [x = 1].

Conclusion: Tangent [y=-1], Normal [x=1].

Question 7

Find tangent and normal to [y = \sqrt{x}] at [x = 9].

Step-by-step solution:

1. Write derivative: [\dfrac{d}{dx}( \sqrt{x}) ][= \dfrac{1}{2\sqrt{x}}].
2. At [x=9]: [m_{t} ][= \dfrac{1}{2\cdot 3} ][= \dfrac{1}{6}].
3. Point: [y(9)=3] → [(9,3)].
4. Tangent: [y – 3 ][= \dfrac{1}{6}(x – 9)] ⇒ [y = \dfrac{1}{6}x + \dfrac{3}{2}].
5. Normal slope: [m_{n} = -6].
6. Normal: [y – 3 = -6(x – 9)] ⇒ [y = -6x + 57].

Conclusion: Tangent [y ][= \dfrac{1}{6}x + \dfrac{3}{2}], Normal [y = -6x + 57].

Question 8

For [y = \tan x], find tangent and normal at [x = \pi/6].

Step-by-step solution:

1. [\dfrac{d}{dx}(\tan x)=\sec^{2}x].
2. At [x=\pi/6]: [m_{t}][=\sec^{2}(\pi/6) ][= \big(\dfrac{2}{\sqrt{3}}\big)^{2} ][= \dfrac{4}{3}].
3. Point: [y(\pi/6)=\tan(\pi/6)][=\dfrac{1}{\sqrt{3}}] → [(\dfrac{\pi}{6},\dfrac{1}{\sqrt{3}})].
4. Tangent: [y – \dfrac{1}{\sqrt{3}} ][= \dfrac{4}{3}\big(x – \dfrac{\pi}{6}\big)].
5. Normal slope: [m_{n} = -\dfrac{3}{4}].
6. Normal: [y – \dfrac{1}{\sqrt{3}} ][= -\dfrac{3}{4}\big(x – \dfrac{\pi}{6}\big)].

Conclusion: Tangent and normal as above.

Question 9

Find tangent and normal to implicit curve [x^{3}+y^{3}=6xy] at point [(1,2)].

Step-by-step solution:

1. Differentiate implicitly: [3x^{2} + 3y^{2}\dfrac{dy}{dx} ][= 6y + 6x\dfrac{dy}{dx}].
2. Collect [\dfrac{dy}{dx}] terms: [3y^{2}\dfrac{dy}{dx} -6x\dfrac{dy}{dx} ][= 6y – 3x^{2}].
3. Factor: [\dfrac{dy}{dx}(3y^{2} – 6x) ][= 6y – 3x^{2}].
4. So [\dfrac{dy}{dx} ][= \dfrac{6y – 3x^{2}}{3y^{2} – 6x} ][= \dfrac{2y – x^{2}}{y^{2} – 2x}].
5. At (1,2): numerator [2(2) – 1 = 3], denominator [4 – 2 = 2] ⇒ [m_{t} = \dfrac{3}{2}].
6. Point (1,2) used directly. Tangent: [y – 2 = \dfrac{3}{2}(x – 1)].
7. Normal slope: [m_{n} ][= -\dfrac{2}{3}]. Normal: [y – 2 = -\dfrac{2}{3}(x – 1)].

Conclusion: Tangent [y – 2 ][= \dfrac{3}{2}(x – 1)], Normal [y – 2 ][= -\dfrac{2}{3}(x – 1)].

Question 10

Physics-style: position vs time curve [s(t)=4t^{2}+2t] treated as [y = s], [x = t]. At [t=2] find the tangent (velocity) and the normal (in t–s plane) equations.

Step-by-step solution:

1. Velocity = slope of tangent = [\dfrac{ds}{dt} = 8t + 2]. At [t=2]: [m_{t} ][= 8(2)+2 = 18].
2. Point: [s(2)=4(4)+4=20] → [(t,s)=(2,20)].
3. Tangent in t–s plane: [s – 20 = 18(t – 2)].
4. Normal slope: [m_{n} ][= -\dfrac{1}{18}].
5. Normal: [s – 20 ][= -\dfrac{1}{18}(t – 2)].

Conclusion: Tangent (velocity line) [s -20 = 18(t -2)]; Normal [s -20 = -\dfrac{1}{18}(t -2)].