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Applications Of Derivatives – Tangents and Normal — Flashcards

Telangana Open School (TOSS) · Class 12 · Mathematics

23 flashcards for Applications Of Derivatives – Tangents and Normal (Telangana Open School (TOSS) Class 12 Mathematics) to test yourself on key terms.

42 questions23 flashcards16 formulas & key relations4 concepts

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23 Flashcards·
Slope of TangentSlope of NormalEquation of TangentEquation of NormalParametric CurvesFormula ApplicationConcept Understanding
Card 1Slope of Tangent

Find the slope of the tangent to the curve $ y = x^3 - 2x $ at $ x = 2 $.

Answer

Step 1: Differentiate $ y = x^3 - 2x $ $$ \frac{dy}{dx} = 3x^2 - 2 $$ Step 2: Substitute $ x = 2 $ $$ \left.\frac{dy}{dx}\right|_{x=2} = 3(2)^2 - 2 = 12 - 2 = 10 $$ Answer: The slope of the tangen…

Card 2Slope of Normal

Find the slope of the normal to the curve $ x^2 + 3y + y^2 = 5 $ at $ (1, 1) $.

Answer

Step 1: Differentiate implicitly w.r.t. $ x $ $$ 2x + 3\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 $$ Step 2: Solve for $ \frac{dy}{dx} $ $$ \frac{dy}{dx}(3 + 2y) = -2x \Rightarrow \frac{dy}{dx} = \frac{-2…

Card 3Equation of Tangent

Find the equation of the tangent to the circle $ x^2 + y^2 = 25 $ at $ (4, 3) $.

Answer

Step 1: Differentiate implicitly $$ 2x + 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y} $$ Step 2: Find slope at $ (4, 3) $ $$ \left.\frac{dy}{dx}\right|_{(4,3)} = -\frac{4}{3} $$ St…

Card 4Equation of Normal

Find the equation of the normal to the curve $ y = x^3 $ at $ (2, 8) $.

Answer

Step 1: Differentiate $$ \frac{dy}{dx} = 3x^2 $$ Step 2: Slope at $ (2, 8) $ $$ \left.\frac{dy}{dx}\right|_{x=2} = 3(4) = 12 $$ Step 3: Slope of normal = negative reciprocal $$ -\frac{1}{12} $$ …

Card 5Parametric Curves

Find the slope of the tangent to the parametric curve $ x = a(\theta - \sin\theta), y = a(1 - \cos\theta) $ at $ \theta = \frac{\pi}{2} $.

Answer

Step 1: Find $ \frac{dx}{d\theta} $ and $ \frac{dy}{d\theta} $ $$ \frac{dx}{d\theta} = a(1 - \cos\theta),\quad \frac{dy}{d\theta} = a\sin\theta $$ Step 2: $ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\th…

Card 6Formula Application

When do you use the formula for the slope of a tangent?

Answer

Use the derivative $ \frac{dy}{dx} $ to find the slope of the tangent when: - The curve is given as $ y = f(x) $ and you need the slope at a point. - The tangent touches the curve at one point and it…

Card 7Equation of Tangent

Apply the formula for the equation of a tangent to $ y = x^2 $ at $ (1, 1) $.

Answer

Step 1: $ y = x^2 \Rightarrow \frac{dy}{dx} = 2x $ Step 2: At $ (1,1) $, slope = $ 2(1) = 2 $ Step 3: Equation using point-slope form: $$ y - 1 = 2(x - 1) \Rightarrow y = 2x - 1 $$ Answer: Tangent…

Card 8Concept Understanding

Why is the normal perpendicular to the tangent?

Answer

The normal is defined as the line perpendicular to the tangent at the point of contact. Since the tangent represents the direction of the curve, the normal points in the direction of maximum change (i…

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Frequently Asked Questions

What are the important topics in Applications Of Derivatives – Tangents and Normal for Telangana Open School (TOSS) Class 12 Mathematics?
Key topics in Applications Of Derivatives – Tangents and Normal include Slope of Tangent and Normal, Equations of Tangent and Normal, Lengths of Tangent, Normal, Subtangent, and Subnormal, Rolle's and Lagrange's Mean Value Theorems. Study these first, then practise questions on each for the Telangana Open School (TOSS) Class 12 board exam.
How many flashcards are available for Applications Of Derivatives – Tangents and Normal?
There are 23 flashcards for Applications Of Derivatives – Tangents and Normal covering key definitions, facts and ideas. A few sample cards are shown on this page.
How should I revise Applications Of Derivatives – Tangents and Normal for the Telangana Open School (TOSS) Class 12 board exam?
Learn the core ideas first, then work through the 42 practice questions on Applications Of Derivatives – Tangents and Normal. Revise definitions regularly and use flashcards for quick recall before the exam.

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