JEE Mathematics Application of Derivatives MCQs Set 30

Answer: (a) \( \frac{27\pi}{8}(2x+1)^2 \)

Explanation:
1. The radius is \( r=\frac34(2x+1) \), so \( V=\frac43\pi r^3=\frac43\pi\cdot\frac{27}{64}(2x+1)^3=\frac{9\pi}{16}(2x+1)^3 \).
2. \( \frac{dV}{dx}=\frac{9\pi}{16}\cdot3(2x+1)^2\cdot2=\frac{27\pi}{8}(2x+1)^2 \).

Teacher's Note:
1. Find the radius first, then differentiate the volume.

Answer: (b) \( \frac{1}{48\pi} \) cm/s

Explanation:
1. \( r=6h \), so \( V=\frac13\pi r^2h=12\pi h^3 \).
2. \( \frac{dV}{dt}=36\pi h^2\frac{dh}{dt}=12 \). At \( h=4 \): \( 576\pi\frac{dh}{dt}=12 \), so \( \frac{dh}{dt}=\frac{1}{48\pi} \) cm/s.

Teacher's Note:
1. Express the volume in terms of h alone.

Answer: (c) Rs. 20.967

Explanation:
1. Marginal cost \( =C'(x)=0.021x^2-0.006x+15 \).
2. At \( x=17 \): \( 0.021\times289-0.102+15=6.069-0.102+15=20.967 \).

Teacher's Note:
1. Marginal cost is the derivative of the cost function.

Answer: (d) \( \sqrt3\,b \) cm²/s

Explanation:
1. With equal sides a: \( A=\frac b4\sqrt{4a^2-b^2} \), so \( \frac{dA}{dt}=\frac b4\cdot\frac{4a}{\sqrt{4a^2-b^2}}\frac{da}{dt}=\frac{ab}{\sqrt{4a^2-b^2}}\frac{da}{dt} \).
2. At \( a=b \) and \( \frac{da}{dt}=-3 \): \( \frac{b^2}{\sqrt3b}(-3)=-\sqrt3\,b \). The area decreases at \( \sqrt3\,b \) cm²/s.

Teacher's Note:
1. Write the area using the height of the isosceles triangle.

Answer: (a) 8 m/s

Explanation:
1. The kite is 150 m above the boy's hand level. If the string length is \( s=250 \) m, then the horizontal distance is \( x=\sqrt{250^2-150^2}=200 \) m.
2. \( s^2=x^2+150^2 \) gives \( s\frac{ds}{dt}=x\frac{dx}{dt} \), so \( \frac{ds}{dt}=\frac{200}{250}\times10=8 \) m/s.

Teacher's Note:
1. The vertical height stays constant, so differentiate \( s^2=x^2+h^2 \).

Answer: (d) \( \frac\pi3 \)

Explanation:
1. \( \frac{d\theta}{dt}=2\frac{d(\sin\theta)}{dt}=2\cos\theta\frac{d\theta}{dt} \).
2. So \( \cos\theta=\frac12 \), giving \( \theta=\frac\pi3 \).

Teacher's Note:
1. Divide by \( \frac{d\theta}{dt} \).

Answer: (c) r

Explanation:
1. \( S=4\pi r^2 \), so \( \frac{dS}{dt}=8\pi r\frac{dr}{dt}=16\pi r \).
2. It is proportional to r.

Teacher's Note:
1. Differentiate with respect to time.

Answer: (c) \( 10\sqrt3 \) cm²/s

Explanation:
1. \( A=\frac{\sqrt3}4s^2 \), so \( \frac{dA}{dt}=\frac{\sqrt3}2s\frac{ds}{dt} \).
2. At \( s=10 \), \( \frac{ds}{dt}=2 \): \( \frac{\sqrt3}{2}\cdot10\cdot2=10\sqrt3 \) cm²/s.

Teacher's Note:
1. Use \( A=\frac{\sqrt3}4s^2 \).

Answer: (d) inversely proportional to the surface area

Explanation:
1. \( \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=k \), so \( \frac{dr}{dt}=\frac{k}{4\pi r^2} \).
2. Since the surface area is \( 4\pi r^2 \), the rate is inversely proportional to the surface area.

Teacher's Note:
1. Note that \( 4\pi r^2 \) is the surface area.

Answer: (c) 8 cm/s

Explanation:
1. They meet when \( 10+6t=3+t^2 \), i.e. \( t^2-6t-7=0 \), so \( t=7 \) s (the positive root).
2. Their speeds are \( 6 \) and \( 2t=14 \) cm/s, so they approach each other at \( |14-6|=8 \) cm/s.

Teacher's Note:
1. Find the time of meeting and the two velocities.

Answer: (a) \( \frac1{30\pi} \) ft/min

Explanation:
1. \( \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=30 \).
2. At \( r=15 \): \( \frac{dr}{dt}=\frac{30}{4\pi\cdot225}=\frac1{30\pi} \) ft/min.

Teacher's Note:
1. Use \( V=\frac43\pi r^3 \).

Answer: (b) \( 3\sqrt3 \)

Explanation:
1. Differentiating \( x^2+y^2=1 \): \( x\frac{dx}{dt}+y\frac{dy}{dt}=0 \).
2. \( \frac12\frac{dx}{dt}=-\frac{\sqrt3}2(-3) \), so \( \frac{dx}{dt}=3\sqrt3 \) (x is increasing).

Teacher's Note:
1. Differentiate the circle equation with respect to time.

Answer: (d) \( 2\sqrt{b^2+c^2} \)

Explanation:
1. \( \frac{d^2x}{dt^2}=-2c \) and \( \frac{d^2y}{dt^2}=2b \).
2. The magnitude of the acceleration is \( \sqrt{4c^2+4b^2}=2\sqrt{b^2+c^2} \).

Teacher's Note:
1. Acceleration is a vector with two components.

Answer: (b) \( 3.92\pi \) m³

Explanation:
1. \( dV=4\pi r^2\,dr=4\pi(49)(0.02) \).
2. \( =3.92\pi \) m³.

Teacher's Note:
1. Use \( dV=V'(r)\,dr \).

Answer: (a) \( -34.995 \)

Explanation:
1. \( f(5)=125-175+15=-35 \) and \( f'(x)=3x^2-14x \), so \( f'(5)=75-70=5 \).
2. \( f(5.001)\approx-35+5(0.001)=-34.995 \).

Teacher's Note:
1. Use \( f(x+h)\approx f(x)+hf'(x) \).

Answer: (a) \( (0,0),(1,2) \) and \( (-1,-2) \)

Explanation:
1. The tangent at x passes through the origin when \( y=xy' \), i.e. \( 4x^3-2x^5=x(12x^2-10x^4) \).
2. This gives \( 8x^5=8x^3 \), so \( x=0,\pm1 \), with points \( (0,0) \), \( (1,2) \) and \( (-1,-2) \).

Teacher's Note:
1. The tangent through the origin satisfies \( y=x\,y' \).

Answer: (b) \( (1,2) \)

Explanation:
1. \( y'=4x-1=3 \) gives \( x=1 \), and \( y=2-1+1=2 \).

Teacher's Note:
1. Equate the slope \( y' \) with 3.

Answer: (c) \( a \)

Explanation:
1. Near \( (-a,a) \) the curve is \( y=-\frac{a^2}{x} \), so \( y'=\frac{a^2}{x^2}=1 \) at \( x=-a \).
2. Subtangent \( =\left|\frac{y}{y'}\right|=a \).

Teacher's Note:
1. Subtangent \( =|y/y'| \).

Answer: (d) \( y^3 \)

Explanation:
1. \( y'=-\frac yx \), so subnormal \( =y\,y'=-\frac{y^2}{x} \).
2. Since \( x=\frac{c^2}{y} \), this is \( -\frac{y^3}{c^2} \), which varies as \( y^3 \).

Teacher's Note:
1. Express the subnormal in a single variable.

Answer: (d) \( x=-1 \) and \( 3 \)

Explanation:
1. \( y'=3x^2-6x-9=3(x-3)(x+1)=0 \).
2. So \( x=3 \) or \( x=-1 \).

Teacher's Note:
1. A horizontal tangent has \( y'=0 \).

Answer: (b) \( \frac12 \)

Explanation:
1. \( \frac{dy}{dx}=\frac{3t^2}{6t}=\frac t2 \).
2. At \( t=1 \) the slope is \( \frac12 \).

Teacher's Note:
1. Use \( \frac{dy/dt}{dx/dt} \).