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For time  𝑡> 0 , the position of an object moving in the xy-plane is given by the parametric equations 𝑥(𝑡) =𝑡𝑐𝑜𝑠(𝑡/2)  and 𝑦(𝑡)=𝑡2+2𝑡 . What is the speed of the object at time 𝑡= 1  ?Responses1.1551.1551.3191.3191.3391.3391.810

Question

For time  𝑡> 0 , the position of an object moving in the xy-plane is given by the parametric equations 𝑥(𝑡) =𝑡𝑐𝑜𝑠(𝑡/2)  and 𝑦(𝑡)=𝑡2+2𝑡 . What is the speed of the object at time 𝑡= 1  ?Responses1.1551.1551.3191.3191.3391.3391.810

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Solution

To find the speed of the object at time t=1, we first need to find the velocity of the object. The velocity is the derivative of the position with respect to time.

First, let's find the derivative of x(t) and y(t) with respect to time t.

The derivative of x(t) = t*cos(t/2) with respect to t is given by the product rule and chain rule of differentiation. The derivative of t is 1 and the derivative of cos(t/2) is -sin(t/2)1/2. So, the derivative of x(t) is dx/dt = cos(t/2) - tsin(t/2)*1/2.

Similarly, the derivative of y(t) = t^2 + 2t with respect to t is dy/dt = 2t + 2.

Now, the speed of the object at time t is given by the magnitude of the velocity vector, which is sqrt[(dx/dt)^2 + (dy/dt)^2].

Substitute t=1 into dx/dt and dy/dt, we get dx/dt = cos(1/2) - 1*sin(1/2)1/2 = 0.877582 - 0.479426 = 0.398156 and dy/dt = 21 + 2 = 4.

So, the speed at t=1 is sqrt[(0.398156)^2 + 4^2] = sqrt[0.158534 + 16] = sqrt[16.158534] = 4.018.

So, the speed of the object at time t=1 is approximately 4.018. This is not one of the options given in the question. Please check the question again.

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