85142
The radius of a circular plate is increasing at the rate of \(0.01 \mathrm{~cm} / \mathrm{s}\) when the radius is \(12 \mathrm{~cm}\). Then, the rate at which the area increases, is
85143
A particle moves along the curve \(y=x^{2}+2 x\). Then, the point on the curve such that \(x\) and \(y\) coordinates of the particle change with the same rate is
1 \((1,3)\)
2 \(\left(\frac{1}{2}, \frac{5}{2}\right)\)
3 \(\left(-\frac{1}{2},-\frac{3}{4}\right)\)
4 \((-1,-1)\)
Explanation:
(C) : Given, \(y=x^{2}+2 x \tag{i}\) \(\because \mathrm{x}\) and \(\mathrm{y}\) coordinates of particle change with same rate. \(\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{dy}}{\mathrm{dt}}\) From eq \({ }^{\mathrm{n}}(\mathrm{i})\), \(y=x^{2}+2 x\) \(\frac{d y}{d t}=2 x \cdot \frac{d x}{d t}+2 \frac{d x}{d t} \Rightarrow \frac{d y}{d t}=(2 x+2) \frac{d x}{d t}\) \(\frac{d y}{d t}=(2 x+2) \frac{d y}{d t}\) \(2 x+2=1\) \(2 x=-1\) \(x=\frac{-1}{2}\) From eq \({ }^{\mathrm{n}}\) (i) \(y=\left(-\frac{1}{2}\right)^{2}+2\left(-\frac{1}{2}\right)\) \(y=\frac{1}{4}-1 \Rightarrow y=-\frac{3}{4}\) Coordinates are, \((\mathrm{x}, \mathrm{y})=\left(-\frac{1}{2}, \frac{-3}{4}\right)\)
AP EAMCET-2004
Application of Derivatives
85144
A point is moving on \(y=4-2 x^{2}\). The \(x-\) coordinate of the point is decreasing at the rate of 5 units/second. Then, the rate at which \(y\) coordinate of the point is changing when the point is at \((1,2)\) is
1 5 unit/s
2 10 unit/s
3 15 unit/s
4 20 unit/s
Explanation:
(D) : Given, \(\frac{\mathrm{dx}}{\mathrm{dt}}=5 \text { unit / sec ond }\) \(\because \quad y=4-2 x^{2}\) \(\frac{\mathrm{dy}}{\mathrm{dt}}=0-4 \mathrm{x} \cdot \frac{\mathrm{dx}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-4 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}\) At point \((1,2)\) \(\frac{\mathrm{dy}}{\mathrm{dt}}=-4 \times 1 \times 5=-20 \Rightarrow\left|\frac{\mathrm{dy}}{\mathrm{dt}}\right|=20 \text { unit } / \mathrm{sec}\)
AP EAMCET-2004
Application of Derivatives
85145
The volume of sphere is increasing at the rate of \(1200 \mathrm{cubic} \mathrm{cm} / \mathrm{s}\). The rate of increase in its surface area when the radius is \(10 \mathrm{~cm}\) is
85142
The radius of a circular plate is increasing at the rate of \(0.01 \mathrm{~cm} / \mathrm{s}\) when the radius is \(12 \mathrm{~cm}\). Then, the rate at which the area increases, is
85143
A particle moves along the curve \(y=x^{2}+2 x\). Then, the point on the curve such that \(x\) and \(y\) coordinates of the particle change with the same rate is
1 \((1,3)\)
2 \(\left(\frac{1}{2}, \frac{5}{2}\right)\)
3 \(\left(-\frac{1}{2},-\frac{3}{4}\right)\)
4 \((-1,-1)\)
Explanation:
(C) : Given, \(y=x^{2}+2 x \tag{i}\) \(\because \mathrm{x}\) and \(\mathrm{y}\) coordinates of particle change with same rate. \(\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{dy}}{\mathrm{dt}}\) From eq \({ }^{\mathrm{n}}(\mathrm{i})\), \(y=x^{2}+2 x\) \(\frac{d y}{d t}=2 x \cdot \frac{d x}{d t}+2 \frac{d x}{d t} \Rightarrow \frac{d y}{d t}=(2 x+2) \frac{d x}{d t}\) \(\frac{d y}{d t}=(2 x+2) \frac{d y}{d t}\) \(2 x+2=1\) \(2 x=-1\) \(x=\frac{-1}{2}\) From eq \({ }^{\mathrm{n}}\) (i) \(y=\left(-\frac{1}{2}\right)^{2}+2\left(-\frac{1}{2}\right)\) \(y=\frac{1}{4}-1 \Rightarrow y=-\frac{3}{4}\) Coordinates are, \((\mathrm{x}, \mathrm{y})=\left(-\frac{1}{2}, \frac{-3}{4}\right)\)
AP EAMCET-2004
Application of Derivatives
85144
A point is moving on \(y=4-2 x^{2}\). The \(x-\) coordinate of the point is decreasing at the rate of 5 units/second. Then, the rate at which \(y\) coordinate of the point is changing when the point is at \((1,2)\) is
1 5 unit/s
2 10 unit/s
3 15 unit/s
4 20 unit/s
Explanation:
(D) : Given, \(\frac{\mathrm{dx}}{\mathrm{dt}}=5 \text { unit / sec ond }\) \(\because \quad y=4-2 x^{2}\) \(\frac{\mathrm{dy}}{\mathrm{dt}}=0-4 \mathrm{x} \cdot \frac{\mathrm{dx}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-4 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}\) At point \((1,2)\) \(\frac{\mathrm{dy}}{\mathrm{dt}}=-4 \times 1 \times 5=-20 \Rightarrow\left|\frac{\mathrm{dy}}{\mathrm{dt}}\right|=20 \text { unit } / \mathrm{sec}\)
AP EAMCET-2004
Application of Derivatives
85145
The volume of sphere is increasing at the rate of \(1200 \mathrm{cubic} \mathrm{cm} / \mathrm{s}\). The rate of increase in its surface area when the radius is \(10 \mathrm{~cm}\) is
85142
The radius of a circular plate is increasing at the rate of \(0.01 \mathrm{~cm} / \mathrm{s}\) when the radius is \(12 \mathrm{~cm}\). Then, the rate at which the area increases, is
85143
A particle moves along the curve \(y=x^{2}+2 x\). Then, the point on the curve such that \(x\) and \(y\) coordinates of the particle change with the same rate is
1 \((1,3)\)
2 \(\left(\frac{1}{2}, \frac{5}{2}\right)\)
3 \(\left(-\frac{1}{2},-\frac{3}{4}\right)\)
4 \((-1,-1)\)
Explanation:
(C) : Given, \(y=x^{2}+2 x \tag{i}\) \(\because \mathrm{x}\) and \(\mathrm{y}\) coordinates of particle change with same rate. \(\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{dy}}{\mathrm{dt}}\) From eq \({ }^{\mathrm{n}}(\mathrm{i})\), \(y=x^{2}+2 x\) \(\frac{d y}{d t}=2 x \cdot \frac{d x}{d t}+2 \frac{d x}{d t} \Rightarrow \frac{d y}{d t}=(2 x+2) \frac{d x}{d t}\) \(\frac{d y}{d t}=(2 x+2) \frac{d y}{d t}\) \(2 x+2=1\) \(2 x=-1\) \(x=\frac{-1}{2}\) From eq \({ }^{\mathrm{n}}\) (i) \(y=\left(-\frac{1}{2}\right)^{2}+2\left(-\frac{1}{2}\right)\) \(y=\frac{1}{4}-1 \Rightarrow y=-\frac{3}{4}\) Coordinates are, \((\mathrm{x}, \mathrm{y})=\left(-\frac{1}{2}, \frac{-3}{4}\right)\)
AP EAMCET-2004
Application of Derivatives
85144
A point is moving on \(y=4-2 x^{2}\). The \(x-\) coordinate of the point is decreasing at the rate of 5 units/second. Then, the rate at which \(y\) coordinate of the point is changing when the point is at \((1,2)\) is
1 5 unit/s
2 10 unit/s
3 15 unit/s
4 20 unit/s
Explanation:
(D) : Given, \(\frac{\mathrm{dx}}{\mathrm{dt}}=5 \text { unit / sec ond }\) \(\because \quad y=4-2 x^{2}\) \(\frac{\mathrm{dy}}{\mathrm{dt}}=0-4 \mathrm{x} \cdot \frac{\mathrm{dx}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-4 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}\) At point \((1,2)\) \(\frac{\mathrm{dy}}{\mathrm{dt}}=-4 \times 1 \times 5=-20 \Rightarrow\left|\frac{\mathrm{dy}}{\mathrm{dt}}\right|=20 \text { unit } / \mathrm{sec}\)
AP EAMCET-2004
Application of Derivatives
85145
The volume of sphere is increasing at the rate of \(1200 \mathrm{cubic} \mathrm{cm} / \mathrm{s}\). The rate of increase in its surface area when the radius is \(10 \mathrm{~cm}\) is
85142
The radius of a circular plate is increasing at the rate of \(0.01 \mathrm{~cm} / \mathrm{s}\) when the radius is \(12 \mathrm{~cm}\). Then, the rate at which the area increases, is
85143
A particle moves along the curve \(y=x^{2}+2 x\). Then, the point on the curve such that \(x\) and \(y\) coordinates of the particle change with the same rate is
1 \((1,3)\)
2 \(\left(\frac{1}{2}, \frac{5}{2}\right)\)
3 \(\left(-\frac{1}{2},-\frac{3}{4}\right)\)
4 \((-1,-1)\)
Explanation:
(C) : Given, \(y=x^{2}+2 x \tag{i}\) \(\because \mathrm{x}\) and \(\mathrm{y}\) coordinates of particle change with same rate. \(\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{dy}}{\mathrm{dt}}\) From eq \({ }^{\mathrm{n}}(\mathrm{i})\), \(y=x^{2}+2 x\) \(\frac{d y}{d t}=2 x \cdot \frac{d x}{d t}+2 \frac{d x}{d t} \Rightarrow \frac{d y}{d t}=(2 x+2) \frac{d x}{d t}\) \(\frac{d y}{d t}=(2 x+2) \frac{d y}{d t}\) \(2 x+2=1\) \(2 x=-1\) \(x=\frac{-1}{2}\) From eq \({ }^{\mathrm{n}}\) (i) \(y=\left(-\frac{1}{2}\right)^{2}+2\left(-\frac{1}{2}\right)\) \(y=\frac{1}{4}-1 \Rightarrow y=-\frac{3}{4}\) Coordinates are, \((\mathrm{x}, \mathrm{y})=\left(-\frac{1}{2}, \frac{-3}{4}\right)\)
AP EAMCET-2004
Application of Derivatives
85144
A point is moving on \(y=4-2 x^{2}\). The \(x-\) coordinate of the point is decreasing at the rate of 5 units/second. Then, the rate at which \(y\) coordinate of the point is changing when the point is at \((1,2)\) is
1 5 unit/s
2 10 unit/s
3 15 unit/s
4 20 unit/s
Explanation:
(D) : Given, \(\frac{\mathrm{dx}}{\mathrm{dt}}=5 \text { unit / sec ond }\) \(\because \quad y=4-2 x^{2}\) \(\frac{\mathrm{dy}}{\mathrm{dt}}=0-4 \mathrm{x} \cdot \frac{\mathrm{dx}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-4 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}\) At point \((1,2)\) \(\frac{\mathrm{dy}}{\mathrm{dt}}=-4 \times 1 \times 5=-20 \Rightarrow\left|\frac{\mathrm{dy}}{\mathrm{dt}}\right|=20 \text { unit } / \mathrm{sec}\)
AP EAMCET-2004
Application of Derivatives
85145
The volume of sphere is increasing at the rate of \(1200 \mathrm{cubic} \mathrm{cm} / \mathrm{s}\). The rate of increase in its surface area when the radius is \(10 \mathrm{~cm}\) is
