What is the Solution to Implicit Differentiation Homework with Given Values?

[jisbon]

Homework Statement

Let $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$
If $\frac{db}{dt}=0.2$ ,$\frac{dc}{dt}=0.3$ , Find $\frac{da}{dt}$ when a=80 , b=100

Relevant Equations

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Homework Statement: Let $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$
If $\frac{db}{dt}=0.2$ ,$\frac{dc}{dt}=0.3$ , Find $\frac{da}{dt}$ when a=80 , b=100
Homework Equations: -

Since we are supposed to find $\frac{da}{dt}$, I can deduce that:
$\frac{da}{dt} =\frac{da}{dc}\cdot\frac{dc}{dt} = (\frac{db}{dc}\cdot\frac{da}{db})\cdot\frac{dc}{dt} =((\frac{db}{dt}\cdot\frac{dt}{dc})\cdot\frac{da}{db})\cdot\frac{dc}{dt} = ((\frac{0.2}{0.3}\cdot\frac{da}{db})\cdot0.3$ = Answer?

To find $\frac{da}{db}$, I need to differentiate implicitly, which gives me $\frac{da}{db}$=1 ? (Not sure if it's correct), hence the answer is 1*0.2/0.3*0.3?

Thanks

 

[andrewkirk]

I need to differentiate implicitly, which gives me $\frac{da}{db}$=1 ? (Not sure if it's correct)

No, that's not correct. Differentiate both sides of the original equation with respect to t, not b. Assume that each of a, b, c is a function of t. You will get an equation with the three derivatives wrt t in it, and the three amounts a, b, c. Making the above numeric substitutions you will get an equation that relates $\frac{da}{dt}$ to c. You can rearrange that to express $\frac{da}{dt}$ in terms of c, which is what I presume they want you to do.

 

[Delta2]

in addition to what @andrewkirk says, you can also find c when a=80,b=100 by using the original equation.

 

[jisbon]

No, that's not correct. Differentiate both sides of the original equation with respect to t, not b. Assume that each of a, b, c is a function of t. You will get an equation with the three derivatives wrt t in it, and the three amounts a, b, c. Making the above numeric substitutions you will get an equation that relates $\frac{da}{dt}$ to c. You can rearrange that to express $\frac{da}{dt}$ in terms of c, which is what I presume they want you to do.

in addition to what @andrewkirk says, you can also find c when a=80,b=100 by using the original equation.

Will it be something like: $\frac{-1}{a^2}\frac{da}{dt}=\frac{-1}{b^2}\frac{db}{dt}+\frac{-1}{c^2}\frac{dc}{dt}$?

 

[Delta2]

yes

 

[jisbon]

Oh ok. Then I guess I could solve the question easily now. Thanks for the clarification

 

[WWGD]

It is often seen as an application of the chain rule as others said. First equation sets up a as a function of b,c and it is assumed both b,c are functions of t. Then we apply the total derivative which uses the chain rule ,i.e, we have a=a(b(t),c(t)) and we go from there to define da/dt.

 

[Mark44]

It is often seen as an application of the chain rule as others said. First equation sets up a as a function of b,c and it is assumed both b,c are functions of t. Then we apply the total derivative which uses the chain rule ,i.e, we have a=a(b(t),c(t)) and we go from there to define da/dt.

Or more simply, without using the total derivative...
Since a is a function of b and c, with b and c being functions of t, then a is also a function of t.
Just differentiate all with respect to t using the chain rule, exactly as @jisbon shows in post #4.

 

[WWGD]

Or more simply, without using the total derivative...
Since a is a function of b and c, with b and c being functions of t, then a is also a function of t.
Just differentiate all with respect to t using the chain rule, exactly as @jisbon shows in post #4.

I don't know of any setup for chain rule using two variables that does not use the total derivative. I have seen, of course, for one variable but not for two or more. But I guess it comes down to the same in the end.

 

[Mark44]

I don't know of any setup for chain rule using two variables that does not use the total derivative. I have seen, of course, for one variable but not for two or more.

You don't need to treat a as a function of two variables; i.e., a = f(b, c).
All you need is to treat a, b, and c as implicit functions of t, and then differentiate implicitly, as hinted at in the title of this thread. IOW, you can interpret the original equation to mean $\frac 1 {a(t)} = \frac 1 {b(t)} + \frac 1 {c(t)}$.

BTW, this is standard fare in first quarter/first semester calculus, usually presented before any discussion of the total derivative.

 

[WWGD]

Well, now the OP knows two techniques and can choose which one to apply. I get to decide since i have 4444 messages, using 44 twice--- not just once;).