Applications Of Derivatives – Tangents and Normal

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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…

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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…

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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…

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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 $$ -\frac{1}{12} $$ …

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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…

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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…

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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…

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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…