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C3 Differentitaion watch

1. Got a question here I am not sure about, would appreciate it if someone could help please:

The diagram shows the curve x=(37+10y-2y^2)^0.5

Part 1 says find an expressio for dx/dy in terms of y

Part 1 says hence find the equation of the tangent to the curve at 7,3

I get (5-2y)/(37-10y-2y^2)^-0.5 to part 1, but when U plug y in to get the gradient it comes out as a negative square root on the bottom, help please?
2. sorry for the typos, obviously I have repeated 'part 1', second should be part 2
3. because it should be 37 + 10y...... in the bracket not minus ....
4. Nevermind!
5. ah brilliant, thought I was missing something obvious! Cheers
6. (Original post by moonaldo99)
Got a question here I am not sure about, would appreciate it if someone could help please:

The diagram shows the curve x=(37+10y-2y^2)^0.5

Part 1 says find an expressio for dx/dy in terms of y

Part 1 says hence find the equation of the tangent to the curve at 7,3

I get (5-2y)/(37-10y-2y^2)^-0.5 to part 1, but when U plug y in to get the gradient it comes out as a negative square root on the bottom, help please?

part 1 seems right

except you've copied out U wrong.

U = 37+10y-2y^2 but you put -10y instead of +10y

Now slot your y=3 in. Voila.

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