Trig word problem

[ogg]

Hello, I am currently studying for a unit final test and i am stumped on a text book question. The question is
Find values of a and b for the curve x^2y+ay^3=b if the point (1,1) is on the curve and the tangent line at (1,1) has the equation 4x+3y=7.

So far i have found out that the slope is -4/3 based on the last equation but that is about as far as i can get.
Any help would be great thanks.

 

[Bob Brown MSEE]

Find values of a and b for the curve x^2y+ay^3=b

Am I correct to interpret this as...
Find values of a and b for the curve (x^2)y+a(y^3)=b


Hint: (1,1) is on the curve.
Can eliminate a or b before using some tricks.
I see this is NOT calculus, only trig tricks, correct?

 

[Bob Brown MSEE]

(x^2)y+a(y^3)=b

for (1,1)
1+a=b

(x^2)y+a(y^3)= (a+1)

a~0.16

What have you been studying in your class that might hint at how to get the value of a?

 

[Bob Brown MSEE]

(x^2)y+a(y^3)=b

for (1,1)
1+a=b

(x^2)y+a(y^3)= (a+1)

a~0.16

What have you been studying in your class that might hint at how to get the value of a?

Have you heard of implicit differentiation?

 

[Bob Brown MSEE]

(x^2)y+a(y^3)= (a+1)

Let's look at point (1,1)
What about x=1+d for very small d?

 

[ogg]

(x^2)y+a(y^3)=b

for (1,1)
1+a=b

(x^2)y+a(y^3)= (a+1)

a~0.16

What have you been studying in your class that might hint at how to get the value of a?

Nothing we have not studyed that yet. But how did u get 0.16 in the first place? missse implict differeation?

i got a as 1/4 and b as 5/4 by equating the slope and the derivative of the function. Still doesnt seem correct tho

 

[HallsofIvy]

Nothing we have not studyed that yet. But how did u get 0.16 in the first place? missse implict differeation?

i got a as 1/4 and b as 5/4 by equating the slope and the derivative of the function. Still doesnt seem correct tho

Yes, if x^2y+ay^3=b then 2xy+ x^2y'+ 3ay^2y'= 0 so that (x^2+ 3ay^2)y'= -2xy. Since you are talking about x= 1, y= 1, then (1+ 3a)y'= -2 or y'= -2/(1+ 3a). As you say, that must be -4/3 so you can solve that for a. I did not get -1/4 for a.

 

[ogg]

Yes, if x^2y+ay^3=b then 2xy+ x^2y'+ 3ay^2y'= 0 so that (x^2+ 3ay^2)y'= -2xy. Since you are talking about x= 1, y= 1, then (1+ 3a)y'= -2 or y'= -2/(1+ 3a). As you say, that must be -4/3 so you can solve that for a. I did not get -1/4 for a.

ahhh i realised why. I messed up the original equation

It is x^2y+ay^2=b
so this would give me -2xy/(x^2+2ya)=Y¹
i than equate that and subsite -4/3 into y1
so i get 1/4 for a.

 

[JeffM]

ahhh i realised why. I messed up the original equation

It is x^2y+ay^2=b
so this would give me -2xy/(x^2+2ya)=Y¹
i than equate that and subsite -4/3 into y1
so i get 1/4 for a.

Looks good to me so it is now trivial to solve the rest of the problem.