Class 12 Maths Chapter 6 Application of Derivatives – Exercise 6.4 NCERT Solutions

Introduction

Exercise 6.4 focuses on tangent and normal to curves using derivatives. Students learn how to find the slope of tangent, equation of tangent, and equation of normal at a given point on a curve. This exercise is crucial for geometry, calculus, and applications in physics.

Formulas Used

1. Slope of Tangent: For curve $y=f(x)$, slope at point $(x_0,y_0)$:

$$m={\frac{dy}{dx}\mid }_{x=x_0}$$

2. Equation of Tangent:

$$y-y_0=m(x-x_0)$$

3. Equation of Normal: Slope of normal = $-\frac{1}{m}$.

$$y-y_0=-\frac{1}{m}(x-x_0)$$

Students Frequently Make Mistakes

• Forgetting to substitute point coordinates into tangent equation.

• Confusing slope of tangent with slope of curve itself.

• Errors in reciprocal slope for normal.

• Skipping derivative evaluation at given point.

• Mixing up tangent and normal equations.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. Find equation of tangent to $y=x^2$ at $(1,1)$.

$$\frac{dy}{dx}=2x,m=2$$

Equation: $y-1=2(x-1)$.

Q2. Find equation of normal to $y=x^2$ at $(1,1)$. Slope of tangent = 2 ⇒ slope of normal = $-\frac{1}{2}$. Equation: $y-1=-\frac{1}{2}(x-1)$.

Q3. Find tangent to $y=\sin x$ at $(\pi \mathrm{/}4,\sin (\pi \mathrm{/}4))$.

$$\frac{dy}{dx}=\cos x,m=\cos (\pi \mathrm{/}4)=\frac{\sqrt{2}}{2}$$

Equation: $y-\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}(x-\frac{\pi }{4})$.

Q4. Find normal to $y=\cos x$ at $(0,1)$. Slope of tangent = $-\sin 0=0$. Normal is vertical line: $x=0$.

Q5. Find tangent to $y=e^x$ at $(0,1)$.

$$\frac{dy}{dx}=e^x,m=1$$

Equation: $y-1=1(x-0)$.

Q6. Find tangent to $y=\ln x$ at $(1,0)$.

$$\frac{dy}{dx}=\frac{1}{x},m=1$$

Equation: $y-0=1(x-1)$.

Q7. Find tangent to $y=x^3$ at $(2,8)$.

$$\frac{dy}{dx}=3x^2,m=12$$

Equation: $y-8=12(x-2)$.

Q8. Find normal to $y=x^3$ at $(2,8)$. Slope of tangent = 12 ⇒ slope of normal = $-\frac{1}{12}$. Equation: $y-8=-\frac{1}{12}(x-2)$.

Q9. Find tangent to $y=\tan x$ at $(\pi \mathrm{/}4,1)$.

$$\frac{dy}{dx}={\sec }^{2}x,m={\sec }^2(\pi \mathrm{/}4)=2$$

Equation: $y-1=2(x-\frac{\pi }{4})$.

Q10. Find tangent to circle $x^2+y^2=25$ at $(3,4)$. Slope of radius = $\frac{4}{3}$. Slope of tangent = $-\frac{3}{4}$. Equation: $y-4=-\frac{3}{4}(x-3)$.

FAQs (10)

FAQ1. What is slope of tangent? Derivative at given point.

FAQ2. What is slope of normal? Negative reciprocal of tangent slope.

FAQ3. How to find tangent equation? Use point‑slope form with derivative.

FAQ4. How to find normal equation? Use reciprocal slope with point‑slope form.

FAQ5. What if tangent slope = 0? Tangent is horizontal line.

FAQ6. What if tangent slope undefined? Tangent is vertical line.

FAQ7. Why tangent important? Represents instantaneous slope of curve.

FAQ8. Why normal important? Perpendicular to tangent, useful in geometry.

FAQ9. Can tangent and normal intersect curve at same point? Yes, always at point of tangency.

FAQ10. Why is Exercise 6.4 important? It builds foundation for geometry and calculus applications.

Conclusion

Exercise 6.4 has 10 solved questions and 10 FAQs that strengthen your understanding of tangent and normal to curves using derivatives. This builds the foundation for geometry and calculus applications in Class 12 Maths.

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