# Calculus Problems And Answers

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**PRE-CALCULUS WORD PROBLEMS, BEST ANSWER GETS 10 POINTs!?**

Ok before you answer this it says to, translate each sentence into a mathematical eqaution, and be sure to identify the meaning of all symbols.

Question #1:A motorboat heads upstream on a river that has a current of 3 MPH. The trip upstream takes 5 hours, while the return trip takes 2.5 hours. What is the speed of the motorboat? (Assume motorboat maintains a constant speed.)

Question #2:An adjustable water sprinkler that sprays water in a circular pattern is placed at the center of a square field whose area is 1250 square feet. What is the shortest radius setting that can be used if the field is to be completely enclosed within the circle?

Thanks.

Let’s make the speed of the motorboat ‘m’.

When the boat travels upstream, its speed will be retarded by the speed of the current – resultant speed m-3. When the boat travels downstream, its speed will be increased by the speed of the current – resultant speed m+3.

Velocity = distance/time {substitute values into the equation}

m-3 = d/5 {for upstream} and m+3 = d/2.5 {for downstream}

This gives us two equations in two unknowns which we can solve simultaneously.

Halve the second equation and we have:

(m+3)/2 = d/5

Subtract this from the first equation and we have:

m – 3 = (m+ 3)/2 {multiply both sides by 2}

2m – 6 = m + 3 {subtract 3 from both sides}

2m – 9 = m {subtract m from both sides}

m – 9 = 0 {Add 9 to both sides}

m = 9

Checking

9-3 = d/5 9+3 = d/2.5

6 = d/5 12 = d/2.5

d=30 d = 30

Therefore, the speed of the boat is 9mph

2) Side of the square is sqrt(1250)

To completely enclose the square field, but have a minimum radius, the radius must be half the diagonal of the square.

Diagonal = sqrt[sqrt(1250)^2)+sqrt(1250)^2] = sqrt(2500) = 50

Therefore, minimum radius for sprinkler to meet given conditions is 50′

**Lec 30 | MIT 18.01 Single Variable Calculus, Fall 2007**