Application of Derivatives
Jammu & Kashmir Board · Class 12 · Mathematics
Flashcards for Application of Derivatives — Jammu & Kashmir Board Class 12 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.
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See them allFind the rate of change of the area of a circle with respect to its radius when r = 5 cm.
Answer
Step 1: Use the area formula A = πr². Step 2: Differentiate with respect to r: dA/dr = 2πr. Step 3: Substitute r = 5 cm: dA/dr = 2π(5) = 10π. Answer: The area changes at the rate of 10π cm² per unit i…
r = 5 cm होने पर एक वृत्त के क्षेत्रफल की उसकी त्रिज्या के सापेक्ष परिवर्तन की दर ज्ञात कीजिए।
Answer
चरण 1: क्षेत्रफल का सूत्र A = πr² का उपयोग कीजिए। चरण 2: r के सापेक्ष अवकलन कीजिए: dA/dr = 2πr. चरण 3: r = 5 cm रखने पर: dA/dr = 2π(5) = 10π. उत्तर: त्रिज्या में 1 इकाई की वृद्धि पर क्षेत्रफल 10π cm² …
एक घन की भुजा की लंबाई x है। यदि उसका आयतन V = x³ और पृष्ठीय क्षेत्रफल S = 6x² है, तथा dV/dt = 9 cm³/s और x = 10 cm है, तो dS/dt ज्ञात कीजिए।
Answer
चरण 1: V = x³ का t के सापेक्ष अवकलन कीजिए। इसलिए, dV/dt = 3x²(dx/dt). चरण 2: दिया है dV/dt = 9, इसलिए 9 = 3x²(dx/dt). चरण 3: dx/dt के लिए हल कीजिए: dx/dt = 9/(3x²) = 3/x². चरण 4: x = 10 cm पर, dx/dt =…
A cube has edge length x. If its volume is V = x³ and surface area is S = 6x², find dS/dt when dV/dt = 9 cm³/s and x = 10 cm.
Answer
Step 1: Differentiate V = x³ with respect to t. So, dV/dt = 3x²(dx/dt). Step 2: Given dV/dt = 9, so 9 = 3x²(dx/dt). Step 3: Solve for dx/dt: dx/dt = 9/(3x²) = 3/x². Step 4: At x = 10 cm, dx/dt = 3/100…
A stone is dropped into a quiet lake and waves form circles. If the radius increases at 4 cm/s and the radius is 10 cm, find how fast the enclosed area increases.
Answer
Step 1: Use A = πr². Step 2: Differentiate with respect to t. So, dA/dt = 2πr(dr/dt). Step 3: Substitute r = 10 cm and dr/dt = 4 cm/s. Then dA/dt = 2π(10)(4) = 80π. Answer: The area increases at 80π c…
एक पत्थर को शांत झील में गिराया जाता है और तरंगें वृत्त बनाती हैं। यदि त्रिज्या 4 cm/s की दर से बढ़ रही है और त्रिज्या 10 cm है, तो घिरा हुआ क्षेत्रफल कितनी तेजी से बढ़ रहा है, ज्ञात कीजिए।
Answer
चरण 1: A = πr² का उपयोग कीजिए। चरण 2: t के सापेक्ष अवकलन कीजिए। इसलिए, dA/dt = 2πr(dr/dt). चरण 3: r = 10 cm और dr/dt = 4 cm/s रखिए। तब dA/dt = 2π(10)(4) = 80π. उत्तर: क्षेत्रफल 80π cm²/s की दर से बढ़त…
एक आयत की लंबाई 3 cm/min की दर से घट रही है और चौड़ाई 2 cm/min की दर से बढ़ रही है। यदि x = 10 cm और y = 6 cm है, तो परिमाप और क्षेत्रफल की परिवर्तन दर ज्ञात कीजिए।
Answer
चरण 1: दिया है dx/dt = -3 cm/min और dy/dt = 2 cm/min. चरण 2: परिमाप P = 2(x + y). अवकलन कीजिए: dP/dt = 2(dx/dt + dy/dt). इसलिए, dP/dt = 2(-3 + 2) = -2 cm/min. चरण 3: क्षेत्रफल A = x·y. अवकलन कीजिए: dA…
The length of a rectangle is decreasing at 3 cm/min and width is increasing at 2 cm/min. If x = 10 cm and y = 6 cm, find the rate of change of perimeter and area.
Answer
Step 1: Given dx/dt = -3 cm/min and dy/dt = 2 cm/min. Step 2: Perimeter P = 2(x + y). Differentiate: dP/dt = 2(dx/dt + dy/dt). So, dP/dt = 2(-3 + 2) = -2 cm/min. Step 3: Area A = x·y. Differentiate: d…
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