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Alina [70]
3 years ago
9

Please help me to see if this is correct

Mathematics
2 answers:
yuradex [85]3 years ago
8 0

Answer:

You are correct

Step-by-step explanation:

sleet_krkn [62]3 years ago
5 0

Answer: yes, you are correct :)

Step-by-step explanation:

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The radius of a right circular cone is increasing at a rate of 1.9 in/s while its height is decreasing at a rate of 2.2 in/s. At
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Answer:

Volume of the cone is increasing at the rate 9916\pi \frac{in^3}{s}.

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Given: The radius of a right circular cone is increasing at a rate of 1.9 in/s while its height is decreasing at a rate of 2.2 in/s.

To find: The rate at which volume of the cone changing when the radius is 134 in. and the height is 136 in.

Solution:

We have,

\frac{dr}{dt} =1.9 \:\text{in/s}, \frac{dh}{dt}=-2.2\:\text{in/s}, r=134 \:\text{in}, h=136\:\text{in}

Now, let V be the volume of the cone.

So, V=\frac{1}{3}\pi r^{2}h

Differentiate with respect to t.

\frac{dv}{dt} =\frac{1}{3}\pi \left [ r^2\frac{dh}{dt}+h\left ( 2r \right )\frac{dr}{dt} \right ]

Now, on substituting the values, we get

\frac{dv}{dt} =\frac{1}{3}\pi\left [ \left ( 134 \right )^2\left ( -2.2 \right )+\left (  136\right )\left ( 2 \right )\left ( 134 \right )\left ( 1.9 \right ) \right ]

\frac{dv}{dt} =\frac{1}{3}\pi\left [  -39503.2+69251.2 \right ]  

\frac{dv}{dt} =\frac{1}{3}\pi\left [ 29748 \right ]

\frac{dv}{dt} =9916\pi \frac{in^3}{s}

Hence, the volume of the cone is increasing at the rate 9916\pi \frac{in^3}{s}.

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