Let F––(t)=t2i–+tj–+k––𝐹_(𝑡)=𝑡2𝑖_+𝑡𝑗_+𝑘_ and G––(t)=i–+tj–+t2k––𝐺_(𝑡)=𝑖_+𝑡𝑗_+𝑡2𝑘_ be two vector functions defined for each t∈R𝑡∈𝑅. (a) The derivative of F––(t)×G––(t)𝐹_(𝑡)×𝐺_(𝑡) with respect to t𝑡, at t=1𝑡=1 is given by

To find the derivative of the cross product F(t) x G(t), we first need to find the cross product of F(t) and G(t).

The cross product of two vectors is given by the determinant of a 3x3 matrix. The first row of the matrix is the unit vectors i, j, k. The second row is the coefficients of the vector F(t), and the third row is the coefficients of the vector G(t).

So, we have:

F(t) x G(t) = det | i j k | | t² t 1 | | 1 t t² |

= i(t² - t) - j(t² - t) + k(t² - t)

= (t² - t)i - (t² - t)j + (t² - t)k

Now, we need to find the derivative of this with respect to t.

d/dt [F(t) x G(t)] = d/dt [(t² - t)i - (t² - t)j + (t² - t)k]

= [2t - 1]i - [2t - 1]j + [2t - 1]k

Finally, we substitute t = 1 to find the derivative at t = 1:

d/dt [F(1) x G(1)] = [21 - 1]i - [21 - 1]j + [2*1 - 1]k

= i - j + k

So, the derivative of F(t) x G(t) with respect to t, at t = 1 is i - j + k.

To find the derivative of the cross product F(t) x G(t), we first need to find the cross product of the two vector functions F(t) and G(t).

The cross product of two vectors is given by the determinant of a 3x3 matrix. The first row of the matrix is the unit vectors i, j, and k. The second row is the coefficients of the vector F(t), and the third row is the coefficients of the vector G(t).

So, the cross product F(t) x G(t) is:

i(t^2 - t) - j(t^2 - 1) + k(t - t^2)

Next, we need to find the derivative of this cross product with respect to t.

The derivative of a vector function is found by taking the derivative of each component of the vector.

So, the derivative of F(t) x G(t) is:

i(2t - 1) - j(2t) + k(1 - 2t)

Finally, we need to evaluate this derivative at t = 1.

Substituting t = 1 into the derivative gives:

i(21 - 1) - j(21) + k(1 - 2*1)

So, the derivative of F(t) x G(t) at t = 1 is:

i(1) - j(2) - k(1)