Derivative as rate measure MCQ Quiz - Objective Question with Answer for Derivative as rate measure - Download Free PDF

Concept:

Rate of Change and Logarithmic Differentiation:

• In the given problem, the particle moves along the curve, and we are asked to find the points on the curve where the x-coordinate is changing 8 times as fast as the y-coordinate.
• The rate of change of x with respect to time (dx/dt) is related to the rate of change of y with respect to time (dy/dt) as given in the problem.
• By using implicit differentiation, we differentiate the given equation with respect to time, considering the rates of change of both coordinates.
• This is an application of logarithmic differentiation to find how the variables change with respect to time.

Calculation:

Given, the equation of the curve is:

6x = y³ + 2

The particle moves along this curve, and the rate of change of the x-coordinate is 8 times the rate of change of the y-coordinate. We are asked to find the points where this condition holds.

Step 1: Differentiate both sides with respect to time (t):

d/dt[6x] = d/dt[y³ + 2]

Using the chain rule, we get:

6(dx/dt) = 3y² (dy/dt)

Step 2: We are given that dx/dt = 8(dy/dt). Substitute this into the equation:

6 × 8(dy/dt) = 3y² (dy/dt)

Step 3: Cancel (dy/dt) from both sides (assuming dy/dt ≠ 0):

48 = 3y²

Step 4: Solve for y:

y² = 48 / 3 = 16

y = ±4

Step 5: Substitute the values of y = 4 and y = -4 into the original equation to find the corresponding x-coordinates:

For y = 4:

6x = 4³ + 2 = 64 + 2 = 66

x = 66 / 6 = 11

For y = -4:

6x = (-4)³ + 2 = -64 + 2 = -62

x = -62 / 6 = -31 / 3

Step 6: The points on the curve are:

(11, 4) and (-31/3, -4)

∴ The points on the curve are (11, 4) and (-31/3, -4), which corresponds to option (3).

Calculation

Given: Volume of a sphere with radius r is \(V = \frac{4}{3}\pi r^3\).

Let D be the diameter. Then \(D = 2r\) or \(r = \frac{D}{2}\).

\(V = \frac{4}{3}\pi \left(\frac{D}{2}\right)^3 = \frac{4}{3}\pi \frac{D^3}{8} = \frac{\pi D^3}{6}\)

Differentiating V with respect to D, we get:

\(\frac{dV}{dD} = \frac{d}{dD} \left(\frac{\pi D^3}{6}\right) = \frac{\pi}{6} \frac{d}{dD} (D^3) = \frac{\pi}{6} (3D^2) = \frac{\pi D^2}{2}\)

Since \(D = 2r\), we have:

\(\frac{dV}{dD} = \frac{\pi (2r)^2}{2} = \frac{\pi (4r^2)}{2} = 2\pi r^2\)

Hence option 2 is correct.

Formula used:

Volume of a cylinder (V) = \(\pi r^2 h\)

where r is the radius and h is the height (depth in this case).

Calculation:

Given:

Diameter of the cylindrical tank (d) = 20 m

Rate of filling the tank \(\frac{dV}{dt}\) = 314 m³/h

Radius of the tank r = d/2 = 20/2 = 10 m

⇒ \(V = \pi (10)^2 h = 100\pi h\)

Differentiate V with respect to time t:

⇒ \(\frac{dV}{dt} = 100\pi \frac{dh}{dt}\)

Substitute the given value of \(\frac{dV}{dt}\)

⇒ \(314 = 100\pi \frac{dh}{dt}\)

\(\pi = 3.14\):

⇒ \(314 = 100 \times 3.14 \times \frac{dh}{dt}\)

⇒ \(314 = 314 \frac{dh}{dt}\)

⇒ \(\frac{dh}{dt} = \frac{314}{314} = 1\) m/h

Hence, option 4 is the correct answer.

Concept Used:

Volume of a sphere: \(V = \frac{4}{3}\pi r^3\) 

Calculation:

Given:

Initial volume of the balloon: \(V_0 = 4500\pi\) cubic meters

Rate of gas leakage: \(72\pi\) cubic meters per minute

Time after leakage began: \(t = 49\) minutes

Volume of gas remaining after 49 minutes:

⇒ \(V = V_0 - 72\pi t = 4500\pi - 72\pi \times 49 = 972\pi\) cubic meters

Radius of the balloon at this time:

⇒ \(V = \frac{4}{3}\pi r^3 = 972\pi\)

⇒ \(r^3 = 729\)

⇒ \(r = 9\) meters

\(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)

We know \(\frac{dV}{dt} = -72\pi\) (negative sign indicates decrease), and \(r = 9\).

⇒ \(-72\pi = 4\pi \times 9^2 \frac{dr}{dt}\)

⇒ \(\frac{dr}{dt} = -\frac{72}{4 \times 81} = -\frac{2}{9}\) meters per minute

∴ The rate at which the radius decreases is \(\frac{2}{9}\) meters per minute.

Hence, option 3 is correct

Concept Used:

Surface area of cube = \(6a^2\)

Surface area of sphere = \(4\pi r^2\)

Calculation

Let side of cube = a

Let radius of sphere = r

Given \(a = 2r\)

Surface area of cube (\(S_c\)) = \(6a^2\)

Surface area of sphere (\(S_s\)) = \(4\pi r^2\)

Differentiate with respect to time (t):

\(\frac{dS_c}{dt} = 12a \frac{da}{dt}\)

\(\frac{dS_s}{dt} = 8\pi r \frac{dr}{dt}\)

Given \(\frac{da}{dt} = \frac{dr}{dt}\)

\(\frac{dS_c}{dt} = 12(2r) \frac{dr}{dt} = 24r \frac{dr}{dt}\)

\(\frac{dS_s}{dt} = 8\pi r \frac{dr}{dt}\)

Ratio of increase of surface area:

\(\frac{dS_c/dt}{dS_s/dt} = \frac{24r \frac{dr}{dt}}{8\pi r \frac{dr}{dt}}\)

\(\frac{dS_c/dt}{dS_s/dt} = \frac{24}{8\pi} = \frac{3}{\pi}\)

∴ The ratio of the increase of their surface area is 3 : π.

Hence option 1 is correct

Concept:

Rate of change of 'x' is given by \(\rm \frac {dx}{dt}\)

Calculation:

Given that, y = 2x – x² and \(\rm \frac {dx}{dt}\) = 3 units/sec

Then, the slope of the curve, \(\rm \frac {dy}{dx}\) = 2 - 2x = m

⇒\(\rm \frac {dm}{dt}\)  = 0 - 2 × \(\rm \frac {dx}{dt}\)

= -2(3)

= -6 units per second

Hence, the slope of the curve is decreasing at the rate of 6 units per second when x is increasing at the rate of 3 units per second.

Hence, option (2) is correct.

CONCEPT:

• If y = f(x), then dy/dx denotes the rate of change of y with respect to x its value at x = a is denoted by: \({\left[ {\frac{{dy}}{{dx}}} \right]_{x = a}}\)
• Decreasing rate is represented by negative sign whereas increasing rate is represented by positive sign.
• Volume of cylinder is given by: πr²h where r is the radius and h is the altitude.

CALCULATION:

Given: dr/dt = 3m/s and dh/dt = - 4m/s

As we know that, volume of cylinder is given by: πr2h where r is the radius and h is the altitude.

Let V = πr2h

Now by differentiating v with respect to t we get,

\(\Rightarrow \frac{{dV}}{{dt}} = 2\pi rh \cdot \frac{{dr}}{{dt}} + \pi {r^2} \cdot \frac{{dh}}{{dt}}\)

Now by substituting r = 4m, h = 6m, dr/dt = 3m/s and dh/dt = - 4m/s we get,

\(\Rightarrow {\left( {\frac{{dV}}{{dt}}} \right)_{\left( {r = 4,\;\;\;h = 6} \right)}} = 2\pi \cdot \left( 4 \right) \cdot \left( 6 \right) \cdot \left( 3 \right) + \pi \cdot {\left( 4 \right)^2} \cdot \left( { - \;4} \right) = 80\pi \)

Hence, option D is the correct answer.

Concept:

• The area of a circle with r units, is: πr² sq. units.
• The rate of change of the value of a function f(x) with respect to a variable t, is given by: \(\rm \frac{d}{dt}f(x)\).

• Chain Rule of Derivatives: \(\rm \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\).

• \(\rm \frac{d}{dx}x^n=nx^{n-1}\).

Calculation:

Let's say that the radius of the circle is r meters and let t be the time in seconds.

It is given that \(\rm \frac{dr}{dt}=-\frac{2}{\pi}\ m/sec\).

Now, Area of the circle: A = πr2 m².

Using the chain rule of derivatives:

\(\rm \frac{dA}{dt}=\frac{dA}{dr}\times \frac{dr}{dt}\)

⇒ \(\rm \frac{dA}{dt}=2\pi r\times\frac{-2}{\pi}\ m^2/sec\)

⇒ \(\rm \frac{dA}{dt}=-4r\ m^2/sec\)

When the radius is 10 m, the rate of change of area will be: -40 m²/sec.

NOTE: A negative value of the rate of change indicates a decrease.

Calculation:

Given Radius R = 12 cm

Volume of the spherical balloon V = \(\rm{4π\over 3}× R^3\)

Rate of change of radius V wrt R =\(\rm{dV\over dR}\)

\(\rm{dV\over dR}\) = \(\rm{d\over dR}\left[{4π\over 3}× R^3\right]\)​​

\(\rm{dV\over dR}\) = \(\rm{4π\over 3}×3R^2\) = 4π R²

As R = 12 cm

\(\rm{dV\over dR}\) = 4π × 144 cm² 

\(\rm{dV\over dR}\) = 576π 

Concept:

The rate of change of the value of a function f(x) with respect to a variable t, is given by: \(\rm \frac {df(x)}{dt}\)

Calculation:

Given side of the cube L = 5cm and \(\rm dL\over dt\) = 2 cm/s

Now volume of the cube V = L³ 

\(\rm dV\over dt\) = \(\rm dV\over dL\) × \(\rm dL\over dt\)

\(\rm dV\over dt\) = \(\rm dL^3\over dL\) × 2

\(\rm dV\over dt\) = 3L² × 2

\(\rm dV\over dt\) = 6 × 5² 

\(\rm dV\over dt\) = 150 cm³/s

CONCEPT :

Let f be a real-valued function which is a composite of two functions u and v; i.e., f = v o u. suppose t = u (x) and if both \(\frac{{dt}}{{dx}}\) and \(\frac{{dv}}{{dt}}\) exist.

Then differentiation of function (f) by chain rule is given by \(\frac{{df}}{{dx}} = {\rm{\;}}\frac{{dv}}{{dt}} \times \frac{{dt}}{{dx}}\)

SOLUTION:

The area of a circle A with radius r is given by A = πr² 

∴ The rate of change of area (A) with respect to time (t) is given by chain rule is - \(\frac{{dA}}{{dt}} = \frac{{d\left( {\pi {r^2}} \right)}}{{dt}} = \frac{{d\left( {\pi {r^2}} \right)}}{{dt}}.\frac{{dr}}{{dt}} = 2\pi r.\frac{{dr}}{{dt}}\)

It is given that : \(\frac{{dr}}{{dt}} = 5\) cm

\(\therefore \frac{{dA}}{{dt}} = 2\pi \left( 8 \right)\left( 5 \right) = 80\pi \)

So when the radius of the circular wave will be 8 cm, the enclosed area will increase at the rate of 80π cm²/s.

Concept:

The rate of change of the value of a function f(x) with respect to a variable t, is given by: \(\rm \frac {df(x)}{dt}\)

Calculation:

Given the side of the cube L = 10cm and \(\rm dL\over dt\) = 4 cm/s

Now volume of the cube V = L3 

\(\rm dV\over dt\) = \(\rm dV\over dL\) × \(\rm dL\over dt\)

\(\rm dV\over dt\) = \(\rm dL^3\over dL\) × 4

\(\rm dV\over dt\) = 3L2 × 4

\(\rm dV\over dt\) = 12 × 102 

\(\rm dV\over dt\) = 1200 cm3/s

Calculation:

Given that the side (L) of the cube changes at the rate = \(\rm{dL\over dt}\)= 2 cm/s

Surface Area of the cube V = 6L²

Rate of change of Surface Area A =\(\rm{dA\over dt}\)= \(\rm{dA\over dL}×{dL\over dt}\)

\(\rm{dA\over dt}\) = \(\rm{d\over dL}\left[6L^2\right]× {dL\over dt}\)​​

\(\rm{dA\over dt}\) = \(\rm 12L × 2\) = 24L

As \(\rm{dA\over dt}\) = 216 cm²/s 

216 = 24 × L

L = \(\rm{216\over 24}\) = 9 cm

Solution:

ex In a + ea In x + ea In a

= \(e^{lna^{x}}+e^{lnx^{a}}+e^{lna^{a}}\)

= a^x + x^a + a^a

Now the first derivative is

\(\frac{d}{dx}(a^{x}+x^{a}+a^{a})\)

= \(\frac{d}{dx}(a^{x})+\frac{d}{dx}(x^{a})+\frac{d}{dx}(a^{a})\)

= a^xln a + ax^a-1 + 0

= axln a + axa-1

Calculation:

Given that the edge (L) of the cube increases with rate = \(\rm{dL\over dt}\)= 4 cm/s

Volume of the cube V = L³

Rate of increase of volume V =\(\rm{dV\over dt}\)= \(\rm{dV\over dL}×{dL\over dt}\)

\(\rm{dV\over dt}\) = \(\rm{d\over dL}\left[L^3\right]× {dL\over dt}\)​​

\(\rm{dV\over dt}\) = \(\rm 3L^2 × 4\) = 12L2 

As L = 20 cm 

\(\rm{dV\over dt}\) = 12 × 20² 

\(\rm{dV\over dt}\) = 12 × 400 = 4800 cm³/s