Answer

Verified

468.3k+ views

Hint: Apply the derivative methods. Rate of decrease of volume is given to us. Use that information.

It’s given that the vertical angle is ${120^ \circ }$ so half of it will be ${60^ \circ }$ . Slant height l is given to us as $3cm$ . Let the height be h and radius be r. The question says, water is dripping out from a conical funnel at the uniform rate of $4c{m^3}/{\text{sec}}$ . In case of a full canonical funnel with water, we say volume is equal to the water present in the funnel. Since, water is decreasing with the uniform rate then we can say $\dfrac{{dV}}{{dt}} = - 4$ . The unit $c{m^3}/{\text{sec}}$ also confirms the fact that it’s nothing but volume. One might think that why the negative sign we are taking? It's because of the fact that the water is decreasing by time. For that decrement, we are using negative signs. Now,

$

\sin {60^ \circ } = \dfrac{r}{l} \Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{r}{l} \Rightarrow r = \dfrac{{\sqrt 3 l}}{2} \\

\cos {60^ \circ } = \dfrac{h}{l} \Rightarrow \dfrac{1}{2} = \dfrac{h}{l} \Rightarrow h = \dfrac{l}{2} \\

$

We know the formula for volume of the cone,

\[

V = \dfrac{1}{3}\pi {r^2}h \\

\Rightarrow V = \dfrac{1}{3}\pi {(\dfrac{{\sqrt 3 l}}{2})^2}\dfrac{l}{2}{\text{ }}[r = \dfrac{{\sqrt 3 l}}{2},h = \dfrac{l}{2}] \\

\Rightarrow V = \dfrac{1}{{{3}}} \times \pi \times \dfrac{{{3}{l^2}}}{4} \times \dfrac{l}{2} \\

\Rightarrow V = \dfrac{1}{8}\pi {l^3} \\

\]

On differentiating we’ll be getting,

$

\dfrac{{dV}}{{dt}} = \dfrac{\pi }{8}3{l^2}\dfrac{{dl}}{{dt}} \\

\Rightarrow - 4 = \dfrac{\pi }{8}3{(3)^2}\dfrac{{dl}}{{dt}}{\text{ }}[\dfrac{{dV}}{{dt}} = - 4c{m^3}/\sec ,l = 3 cm] \\

\Rightarrow \dfrac{{dl}}{{dt}} = \dfrac{{ - 4 \times 8}}{{27\pi }} \\

\Rightarrow \dfrac{{dl}}{{dt}} = \dfrac{{ - 32}}{{27\pi }} \\

$

Hence, the rate of the decrease of the slant height of the water is $\dfrac{{ - 32}}{{27\pi }}$ .

Note: The toughest part in this question is to recognise the concept. Once you get it as the question of rates then half of the question is finished there itself. After that, we need to find the rate given to us and then use its formula. It’ll lead us to the solution.

It’s given that the vertical angle is ${120^ \circ }$ so half of it will be ${60^ \circ }$ . Slant height l is given to us as $3cm$ . Let the height be h and radius be r. The question says, water is dripping out from a conical funnel at the uniform rate of $4c{m^3}/{\text{sec}}$ . In case of a full canonical funnel with water, we say volume is equal to the water present in the funnel. Since, water is decreasing with the uniform rate then we can say $\dfrac{{dV}}{{dt}} = - 4$ . The unit $c{m^3}/{\text{sec}}$ also confirms the fact that it’s nothing but volume. One might think that why the negative sign we are taking? It's because of the fact that the water is decreasing by time. For that decrement, we are using negative signs. Now,

$

\sin {60^ \circ } = \dfrac{r}{l} \Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{r}{l} \Rightarrow r = \dfrac{{\sqrt 3 l}}{2} \\

\cos {60^ \circ } = \dfrac{h}{l} \Rightarrow \dfrac{1}{2} = \dfrac{h}{l} \Rightarrow h = \dfrac{l}{2} \\

$

We know the formula for volume of the cone,

\[

V = \dfrac{1}{3}\pi {r^2}h \\

\Rightarrow V = \dfrac{1}{3}\pi {(\dfrac{{\sqrt 3 l}}{2})^2}\dfrac{l}{2}{\text{ }}[r = \dfrac{{\sqrt 3 l}}{2},h = \dfrac{l}{2}] \\

\Rightarrow V = \dfrac{1}{{{3}}} \times \pi \times \dfrac{{{3}{l^2}}}{4} \times \dfrac{l}{2} \\

\Rightarrow V = \dfrac{1}{8}\pi {l^3} \\

\]

On differentiating we’ll be getting,

$

\dfrac{{dV}}{{dt}} = \dfrac{\pi }{8}3{l^2}\dfrac{{dl}}{{dt}} \\

\Rightarrow - 4 = \dfrac{\pi }{8}3{(3)^2}\dfrac{{dl}}{{dt}}{\text{ }}[\dfrac{{dV}}{{dt}} = - 4c{m^3}/\sec ,l = 3 cm] \\

\Rightarrow \dfrac{{dl}}{{dt}} = \dfrac{{ - 4 \times 8}}{{27\pi }} \\

\Rightarrow \dfrac{{dl}}{{dt}} = \dfrac{{ - 32}}{{27\pi }} \\

$

Hence, the rate of the decrease of the slant height of the water is $\dfrac{{ - 32}}{{27\pi }}$ .

Note: The toughest part in this question is to recognise the concept. Once you get it as the question of rates then half of the question is finished there itself. After that, we need to find the rate given to us and then use its formula. It’ll lead us to the solution.

Recently Updated Pages

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Advantages and disadvantages of science

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

10 examples of evaporation in daily life with explanations

Difference Between Plant Cell and Animal Cell

Write a letter to the principal requesting him to grant class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE