Academic Detention...again #1

On the non-calculator portion of our recent AP exam, I found that many people did not know how to work number 23.

Question 23

This question asks for you to approximate the value of y at x=3.1, given that the curve x^3+xtan(y)=27 through (3,0).

1. Implicit differentiation.

x^3+xtan(y)=27
3x^2 + xsec^2(y)dy + tan(y) = 0

2. Separate dy from the rest of your function.

3x^2+xsec^2(y)dy+tan(y)=0
xsec^2(y)dy=(-3x^2)-(tan(y))
dy=((-3x^2)-(tan(y)))/(x(sec^2(y)))

3. Plug in your x and y values.

dy=((-3x^2)-(tan(y)))/(x(sec^2(y)))
dy=((-3(3^2))-(tan(0)))/(3(sec^2(0)))
dy=((-3(9))-(0))/(3(1))
dy=(-27)/3

dy=-9

4. Create your equation for the tangent line.

(y-0)=(-9)(x-3)

5. Plug in 3.1

(y-0)=(-9)(x-3)
y=(-9)((3.1)-3)
y=(-9)(.1)

y=(-0.9)

And there it is. The answer for number 23 on the non-calculator portion of our AP exam.