If a cylindrical tank holds 100,000 gallons of water, which...

If a cylindrical tank holds 100,000 gallons of water, which can be drained from the bottom of the tank in an hour, then Torricelli's Law gives the volume $V$ of water remaining in the tank after $t$ minutes as $$ V(t)=100,000\left(1-\frac{1}{60} t\right)^{2} \quad 0 \leqslant t \leqslant 60 $$ Find the rate at which the water is flowing out of the tank (the instantaneous rate of change of $V$ with respect to $t$ ) as a a function of $t .$ What are its units? For times $t=0,10,20,30$ $40,50,$ and 60 min, find the flow rate and the amount of water remaining in the tank. Summarize your findings in a sentence or two. At what time is the flow rate the greatest? The least?

If a cylindrical tank holds 100,000 gallons of water, which can be drained from the bottom of the tank in an hour, then Torricelli's Law gives the volume $V$ of water remaining in the tank after $t$ minutes as
$$
V(t)=100,000\left(1-\frac{1}{60} t\right)^{2} \quad 0 \leqslant t \leqslant 60
$$
Find the rate at which the water is flowing out of the tank (the instantaneous rate of change of $V$ with respect to $t$ ) as a a function of $t .$ What are its units? For times $t=0,10,20,30$ $40,50,$ and 60 min, find the flow rate and the amount of water remaining in the tank. Summarize your findings in a sentence or two. At what time is the flow rate the greatest? The least?

Key Concepts

Instantaneous Rate of Change

This concept involves finding the derivative of a function, which represents how a quantity changes at a specific moment. In physical contexts, such as fluid dynamics, the instantaneous rate of change can tell us how quickly a quantity like volume is decreasing or increasing at a given time.

Chain Rule in Differentiation

The chain rule is a fundamental tool used for differentiating composite functions, where one function is applied inside another. When the variable inside the outer function is itself a function of time, the chain rule helps in computing the overall derivative correctly.

Rate Units

Understanding the units associated with rates is essential in linking mathematical results with physical interpretations. For example, when differentiating a volume with respect to time, the resulting units often represent a flow rate, such as gallons per minute, which describes the speed of change.

Physical Interpretation of Derivatives

In applied problems, the derivative is more than a mathematical tool—it represents real-world quantities. A negative derivative, for instance, indicates a decrease in the measured quantity over time, such as a draining tank losing water.

Function Evaluation at Specific Times

Evaluating a function and its derivative at particular time points allows one to understand how system behavior evolves over time. This process helps in making concrete predictions and understanding time-dependent trends in dynamic processes.

Transcript

00:01 Instantaneous rate of change of the equation v of t equals 100 ,000 times 1 minus 1 over 60 t squared.

00:07 And to do that, we'll use the equation v prime of a equals limit as x approach as a of f of x minus a of f of a over x minus a.

00:14 And for this example, we are finding the general case.

00:18 So just v prime of t, not v prime of any specific number.

00:22 So we will just use a.

00:24 So to start off, we will have v prime of a equals the limit.

00:30 As x approaches a of 100 ,000 times 1 minus 1 over 60 t squared.

00:41 And i apologize, this sex under the limit should actually be t because t is the variable that we're using in this equation.

00:48 So that will be that minus 100 ,000 times 1 minus 1 over 60 a squared.

00:56 We just replaced t with a all over t minus a.

01:02 So if we can take out a 100 ,000 here, and since it's just a constant, we can take it out of the limit altogether.

01:11 So you're left with 1 minus 1 over 60t squared, minus 1 over 60a squared, divided by t minus a.

01:26 And when we factor that, you will have 100 ,000 times the limit as t burgess a of 1 minus 2 over 60t plus 1 over 60 squared t minus 1 minus 2 over 60a plus 1 over 60 squared a plus 1 over 60 squared a squared and the c is also supposed to be squared.

01:58 And all of that divided by t minus a.

02:02 So to simplify this one and this one cancel up because that's minus 1.

02:07 So we'll have 100 ,000 times the limit as t approaches a of negative 1 over 30t plus 1 over 3 ,600 t squared minus 1 over 30a minus 1 over 3 ,600a squared.

02:33 And all that divided by t minus a.

02:37 So now we'll multiply everything by 3 ,600.

02:41 Sorry, divide everything by 3 ,600.

02:44 So we'll have 100 ,000 divided by 3 ,600 times the limit as t approaches a, of negative 120t plus t squared plus 120a minus a squared, divided by t minus a.

03:08 So now 100 ,000 over 3 ,600 comes out to 250 over 9 times the limit as t approaches a.

03:18 And here we'll take out a negative 120.

03:21 So we're left with t minus a plus t squared minus a divided by t minus a...

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