Example Question - rate of change
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Calculate Rate of Change in Distance with Respect to Time
<p>Let s(t) be the distance travelled as a function of time t.</p> <p>The rate of change of distance with respect to time is the derivative of s with respect to t, which is the velocity v(t) = \frac{ds}{dt}.</p> <p>For constant velocity, the rate of change of distance with respect to time is constant.</p> <p>The rate of change over the 80 km travelled is simply the constant velocity, which can be calculated as v = \frac{\Delta s}{\Delta t} where \Delta s = 80 \text{ km} and \Delta t = t_f - t_i, the time taken to travel the last 80 km.</p> <p>Without additional specific information about the time interval \Delta t or the actual function s(t) for the motion, we cannot compute a numerical value for the rate of change.</p>
Rate of Change in Distance Over Time for Vehicles
<p>Given that the train travels 150 km in 3 hours, the rate of change in distance with respect to time can be calculated as follows:</p> <p>The rate of change is the velocity of the train in km/min, which is constant since the question doesn't provide any information about the acceleration or deceleration of the train.</p> <p>First, convert the time from hours to minutes: \(3 \text{ hours} = 3 \times 60 \text{ minutes} = 180 \text{ minutes}.\)</p> <p>Then calculate the velocity of the train: \(\text{Velocity} = \frac{\text{Distance}}{\text{Time}} = \frac{150 \text{ km}}{180 \text{ minutes}} = \frac{5}{6} \text{ km/min}.\)</p> <p>The train travels at a constant velocity of \( \frac{5}{6} \text{ km/min}\) for the first 42 minutes.</p> <p>Thus, the rate of change of distance with respect to time, \(v\), for the first 42 minutes is \(\frac{5}{6} \text{ km/min}\).</p>
Rate of Change of Triangle Area with Changing Sides and Angle
To solve the given problem, we must determine the rate of change of the area of a triangle with sides A and B and angle B between them, given how A and B are changing with time. We are given: - The area of a triangle is given by the formula \( A = \frac{1}{2} ab \sin{\theta} \), where \( a \) and \( b \) are the sides of the triangle and \( \theta \) is the angle between them. For our problem, we can rewrite the given area formula as \( Area = \frac{1}{2} AB \sin{B} \). - Side \( A \) is increasing at 0.1 units/s. - Side \( B \) is decreasing at 0.1 units/s. - Angle \( B \) is increasing at 0.2 radians/s. We want to find the rate of change of the triangle's area, which can be found by differentiating the area formula with respect to time \( t \). This is an application of the chain rule in calculus. So we differentiate \( Area = \frac{1}{2} AB \sin{B} \) with respect to \( t \): \( \frac{d(Area)}{dt} = \frac{1}{2} \left[ B \sin{B} \frac{dA}{dt} + A \sin{B} \frac{dB}{dt} + AB \cos{B} \frac{dB}{dt} \right] \) Plugging in the rates given: \( \frac{dA}{dt} = 0.1 \, \text{units/s} \) \( \frac{dB}{dt} = -0.1 \, \text{units/s} \) \( \frac{dB}{dt} = 0.2 \, \text{radians/s} \) Substitute these into the derivative: \( \frac{d(Area)}{dt} = \frac{1}{2} \left[ B \cdot \sin{B} \cdot 0.1 + A \cdot \sin{B} \cdot (-0.1) + AB \cdot \cos{B} \cdot 0.2 \right] \) Unfortunately, the image doesn't provide the exact values of \( A \), \( B \), and angle \( B \) at the moment of calculation, so we cannot give a numerical answer. You would need to substitute the current values of \( A \), \( B \), and angle \( B \) into the derivative to find the rate of change at that particular moment.
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