Question

The current I in a device is related to time t by the equation I=10t-5t^2+2.Calculate the maximum current and time taken to reach maximum current.

Answer 1

Explanation:

The current I in a device is related to time t by the equation

I = 10t - 5t² + 2

The equation is a quadratic equation. The plot of this equation on a graph would give a parabola whose vertex would be equal to the maximum current in the device

The vertex of the parabola is calculated as follows,

Vertex = -b/2a

From the equation,

a = - 5

b = 10

Vertex = - 10/2 × - 5 = - 10/ - 10 = 1

Therefore, the time taken to reach maximum current is 1

The maximum current would be

I = (10 × 1)- (5 × 1²) + 2

I = 10- 5 + 2 = 7

Answer 2

• maximum current: 7
• time to reach maximum: 1

Explanation:

The equation can be rewritten to vertex form to find the maximum value:

I = -5(t^2 -2t) +2 . . . . . . factor leading coefficient from variable terms

I = -5(t^2 -2t +1) +2 -(-5(1)) . . . . . add and subtract the square of half the t coefficient

I = -5(t -1)^2 +7 . . . . . . write in vertex form

The vertex of the graph is at (t, I) = (1, 7).

The maximum current of 7 is reached at t=1.