Question

QUESTION 1 [13 MARKS (a) Fill in the blanks with appropriate answer 1 (i) The domain of the function $f(x,y) = \frac{1}{\sqrt{25-x-y}}$ is and the range is (ii) In the equation $a = 3T + 2N$, the normal component of acceleration is equal to and the tangential component of acceleration is equal to . [4 Marks] (b) Given the vector valued function $r(t) = 3\cos(2t)i + 3\sin(2t)j + 3tk$. Find the following: (i) unit tangent vector $T(t)$, (ii) principal unit normal $N(t)$, (iii) curvature at $t = 0$. [9 Marks]

          QUESTION 1 [13 MARKS
(a) Fill in the blanks with appropriate answer
1
(i) The domain of the function $f(x,y) = \frac{1}{\sqrt{25-x-y}}$ is ______ and
the range is ______
(ii) In the equation $a = 3T + 2N$, the normal component of acceleration is equal to ______ and the tangential component of acceleration is equal to ______.
[4 Marks]
(b) Given the vector valued function $r(t) = 3\cos(2t)i + 3\sin(2t)j + 3tk$. Find the following:
(i) unit tangent vector $T(t)$, 
(ii) principal unit normal $N(t)$,
(iii) curvature at $t = 0$.
[9 Marks]

QUESTION 1 [13 MARKS
(a) Fill in the blanks with appropriate answer
1
(i) The domain of the function f(x,y) = (1)/(√(25-x-y)) is and
the range is
(ii) In the equation a = 3T + 2N, the normal component of acceleration is equal to and the tangential component of acceleration is equal to .
[4 Marks]
(b) Given the vector valued function r(t) = 3cos(2t)i + 3sin(2t)j + 3tk. Find the following:
(i) unit tangent vector T(t),
(ii) principal unit normal N(t),
(iii) curvature at t = 0.
[9 Marks]

Added by Juan Antonio N.

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