Answer to Question #180670 in Calculus for Harry

Question

The power (in watts) from an engine is represented by the equation: 𝑃=100𝑡1.4+6𝑡 where t is the time in seconds. i. Draw the graph that represents power against time for this engine. ii. Show the area on the graph which represents the energy converted between 5 s and 15 s. iii. Show, using summation, how you would gain an approximation for the energy. iv. Using integration, evaluate the exact energy between 5 s and 15 s.

Answer 1

Solution

i,ii. Graph for power P(t) = 100𝑡^1.4+6𝑡 with the area on the graph which represents the energy converted between 5s and 15s are shown below

iii. If ∆t – some small time subinterval, the value of this area may be calculated approximately as

$S=\sum_{n=1}^NP(t_n)\Delta t$

where N – number of subintervals of [5,15], t_n = 5+n*∆t, ∆t = (15-5)/N.

The value of this sum is S = 29900.85 for N = 5 or S = 28094.41  for N = 10.

iv. The exact energy between 5s and 15s is the integral E = ∫₅¹⁵P(t)dt = ∫₅¹⁵(100𝑡^1.4+6𝑡)dt = (100𝑡^2.4/2.4+3𝑡²)|₅¹⁵ = 28370.41-2057.97 = 26312.44