Math Help - Help with Slope of the Tangent Line and Related Rates :(

1. Help with Slope of the Tangent Line and Related Rates :(

Guys, we have discussed the slope of the tangent line and also the related rates yesterday and I'am in a trouble..I can't even understand these topics even though my Professor gave already a bunch of example.. I tried solving problems related to these topics and It's hard for me to get the correct answer..
Can you answer these problems which I found on my book and "Please" tell me how you got the answer step by step..Pls.. Thanks in advance guys!

*Slope of the Tangent Line
Find the slope of the tangent line and write an equation of the tangent line to the graph of the equation at the given point P.
x^2y+sin y=2pi at P(1,2pi)

*Related Rates
a.) The Area(A) of an equilateral triangle is decreasing at a rate(dA/dt) of 4cm^2/min. Find the rate(dl/dt) at which the length(l) is charging when its area is 200cm^2
b.) Gas is escaping from a spherical balloon at a rate dV/dt=10ft^3/hr.At what rate(dr/dt) is the radius(r) changing when the volume V=400ft^3

2. Re: Help with Slope of the Tangent Line and Related Rates :(

For the slope, you'll need to do implicit differentiation using the chain rule. Are you aware of how to do that?

For the related rates, we are told that $\frac{dA}{dt}=-4$
I'm not certain at first glance whether it should be negative but I'm going to go with that for now. I'll think about it and correct in a few mins if I decide it should be something else.

We can deduce that the height of the triangle will be $\frac{\sqrt{3}}{2}l$ using Pythagoras. The area of the triangle will be $\frac{l}{2}\times~\frac{\sqrt{3}}{2}l$ using the fact that 'area is half base x height'

So $A=\frac{\sqrt{3}}{4}l^2$

This means that we can find $\frac{dl}{dA}$

Then use the chain rule:

$\frac{dl}{dt}=\frac{dl}{dA}\times \frac{dA}{dt}$

I'll leave part b) for now until we've covered what's here so far.

Hope it helped.