Question

Consider an object moving along the parametrized curve with equations: x(t)=et + e-t, y(t)=e-t where t is in the time interval [0,1] seconds. The maximum speed of the object on the time interval is at time The minimum speed of the object on the time interval is at time

Answer 1

Answer 2

Answer 3

The object is moving in a 2-dimensional space and its position at time 't' is given by:

2-dimensional space: XY plane

X-position: X(t) = e^t + e^-t

Y-position: Y(t) = e^-t

The speed along these axes (X and Y) is given by the first differential of the position function with respect to time.

Hence,

Speed along X-axis:

$V_{x}(t)=\frac{\mathrm{d} X(t)}{\mathrm{d} t}$

Speed along Y-axis:

$V_{y}(t)=\frac{\mathrm{d} Y(t)}{\mathrm{d} t}$

Hence,

V_x(t) = e^t - e^-t

V_y(t) = -e^-t

Total speed of the object is given by:

$V(t)=\sqrt{V_{x}^{2}(t)+V_{y}^{2}(t)}$

Hence,

$V(t)=e^{2t}+2e^{-2t}-2$

The first differential is given by:

$V'(t)=2e^{2t}-4e^{-2t}$

The first differential is zero when:

$2e^{2t}=4e^{-2t} \or\ t=\frac{ln \2}{4}=0.173$

Therefore, V(t) has a minima or maxima at t = ln(2) / 4 = 0.173

To check if it is minima or maxima, we calculate the second differential.

$V''(t)=4e^{2t}+8e^{-2t}$

We notice that the second differential is positive at t = 0.173. This implies that t = 0.173 is a local minima.

Therefore, the speed function decreases from t = 0 to t = 0.173, achieves a local minima at t = 0.173 and then increases till t = 1.

Maximum speed is at t = 0 or t = 1.

V(0) = 1 and V(1) = 5.659

Hence, maximum speed is 5.659 at t = 1

Minimum speed is:

$V(\frac{ln\2}{4})=2\sqrt{2}-2=0.828$

ANSWER:

a) Max. speed is 5.659 at time 1

b) Min. speed is 0.828 at time 0.173

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