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.
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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…
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} $ $$ \frac{dy}{dx}(3 + 2y) = -2x \Rightarrow \frac{dy}{dx} = \frac{-2…
Find the equation of the tangent to the circle $ x^2 + y^2 = 25 $ at $ (4, 3) $.
Answer
Step 1: Differentiate implicitly $$ 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} $$ St…
Find the equation of the normal to the curve $ y = x^3 $ at $ (2, 8) $.
Answer
Step 1: Differentiate $$ \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} $$ …
Find the slope of the tangent to the parametric curve $ x = a(\theta - \sin\theta), y = a(1 - \cos\theta) $ at $ \theta = \frac{\pi}{2} $.
Answer
Step 1: Find $ \frac{dx}{d\theta} $ and $ \frac{dy}{d\theta} $ $$ \frac{dx}{d\theta} = a(1 - \cos\theta),\quad \frac{dy}{d\theta} = a\sin\theta $$ Step 2: $ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\th…
When do you use the formula for the slope of a tangent?
Answer
Use the derivative $ \frac{dy}{dx} $ to find the slope of the tangent when: - The curve is given as $ y = f(x) $ and you need the slope at a point. - The tangent touches the curve at one point and it…
Apply the formula for the equation of a tangent to $ y = x^2 $ at $ (1, 1) $.
Answer
Step 1: $ y = x^2 \Rightarrow \frac{dy}{dx} = 2x $ Step 2: At $ (1,1) $, slope = $ 2(1) = 2 $ Step 3: Equation using point-slope form: $$ y - 1 = 2(x - 1) \Rightarrow y = 2x - 1 $$ Answer: Tangent…
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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