If
A = 5t² i + tj -t³k and B =1/sinti-costi then find d/dt(AB) and d/dt (A×B)
Answers
The derivative of A cross B with respect to t is (10ti + j - 3t²k) × (1/sinti-costi) + (5t² i + tj -t³k) × (-(costi + sinti)/sint)
To find d/dt(AB), we can use the product rule of differentiation, which states that:
d/dt(AB) = d/dt(A)B + Ad/dt(B)
First, let's find d/dt(A). We have:
A = 5t² i + tj -t³k
d/dt(A) = d/dt(5t² i + tj -t³k)
= 10ti + j - 3t²k
Next, let's find d/dt(B). We have:
B = 1/sinti-costi
d/dt(B) = -1/sintcos t di - 1/sint sin t di
= -(cost/sint)di - (sint/sint)di
= -costi/sint - sinti/sint
= -(costi + sinti)/sint
Now, we can substitute the values of d/dt(A) and d/dt(B) into the product rule to find d/dt(AB):
d/dt(AB) = d/dt(A)B + Ad/dt(B)
= (10ti + j - 3t²k) (1/sinti-costi) + (5t² i + tj -t³k) (-(costi + sinti)/sint)
This is the derivative of AB with respect to t.
For d/dt (A×B),
A×B = (5t² i + tj -t³k) × (1/sinti-costi)
d/dt (A×B) = d/dt((5t² i + tj -t³k) × (1/sinti-costi))
= (10ti + j - 3t²k) × (1/sinti-costi) + (5t² i + tj -t³k) × (-(costi + sinti)/sint)
This is the derivative of A cross B with respect to t.
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