# Tutor profile: Isaias T.

## Questions

### Subject: Pre-Calculus

Convert the polar coordinate ( 4 , \frac{3}{4\pi} )

We know that the relationships we should know for converting coordinates are x=rcos(\theta) and y=rsin(\theta). From here we are able to plug into the equations and find that the answer is ( -2\sqrt{2} , 2\sqrt{2} ).

### Subject: Calculus

A spherical balloon is inflated so that its Volume is increasing at a rate of 15 in^3/min. How fast is the Surface Area of the balloon increasing when the radius is 42 in?

We know that Volume of a sphere is equal to \frac{4}{3} \pi r^3 and the Surface Area is equal to 4 \pi r^2. Because we are missing information for the Surface Area we must work with the Volume first. From here we use implicit differentiation to get the derivative of Volume with respect to time and get the surface area equation multiplied by the rate of change of the radius. Now we plug in the given radius and solve for the rate of change of the radius and find it to be \frac{5}{7056\pi}. Now we get the derivative of Surface Area with respect to time and find that it is the rate of change of the radius multiplied by four times the circumference of a circle with radius r. Finally, we plug in the radius and the rate of change for this radius to find the rate of change for the Surface Area. The rate at which the surface area is increasing is 5/7 in^2/min or 0.714285 in^2/min.

### Subject: Algebra

A rectangular animal exhibit is enclosed by 524 feet of fencing. If the length of the exhibit is two less than five times the width, what is the length in yards?

First we write the relationship that we are given and that is that \ell = 5w-2. From here we are able to write down the perimeter equation in terms of w: 2(5w-2)+2w = 524. Then we must solve for w and we get that w = 44 ft. Next, we plug the w into our equation for \ell and get 218 ft. Finally we convert this to yards by multiplying by yards per feet. The length of the animal exhibit is 72 yards.

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