MATH113 Calculus and Geometry
Test 2016: Solutions
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1.
i) . [2]
. [2]
ii) The point is . [2]
iii) The area of is . [1] We have . [2] The length of this vector is . [1]
iv) We have [1] and . [1]
Further, . [1]
Thus, , and hence . [1]
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2.
i) To calculate the arc length, we recall the formula: . [1]
Here [1], hence . [1] This equals . [1] It follows that the arc length is . [1]
ii) We have [1] and [1]. So an equation of the tangent line to the image of at the point is . [1]
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3.
The surface is , where . [1]
We have [1] and . [1]
So the normal line is [1]
and the tangent plane is . [2]
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4.
i) We check the partial derivatives ([1] for knowing the right condition). We have: and [1], so the partial derivatives do not match and hence cannot be expressed as . [1]
ii) Here , . [1] Similarly , . [1] Finally, , [1], so all of the conditions are satisfied and hence can be expressed as for some . To find , since , we obtain: . [1] Now, checking against , , we obtain:
Hence we have: and , so we deduce: (). It follows that . [1]
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5.
Let . We form the auxiliary function . [2] Then we look for stationary points:
By the first equation, we have or . [1]
If , then (since ) and hence .
If , then (by the second equation) and hence .
So the possible extreme points are . [1]
We have , , , .
We deduce that the greatest value of is 2 [1], and the least value is . [1]
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6.
The region is given by: [2], so
Now
so . [1]