# Thread: Implicit Derivatives?

1. ## Implicit Derivatives?

I was wondering how this involves using implicit differentiation?

v = v(t)
a= a(t)
v=sqrt(8s+16)

since v(t) = sqrt(8s+16) do i have to use implicit differentiation? How would finding it help me show that acceleration is constant?

Also, for Part C, do i just graph v(t) = sqrt(8s+16)

I'm lost in how i can use implicit differentiation to solve this.

Snapshot: https://imgur.com/a/9Cs9q

2. ## Re: Implicit Derivatives?

What is the question? You define the functions v and a but don't say what it is you want to do with them! And you say "Also, for Part C, do i just graph v(t) = sqrt(8s+16)" but don't say what "part C" says! How could we possibly say know what you are to do if we don't know what the question is?

3. ## Re: Implicit Derivatives?

(a) $a=\dfrac{dv}{dt} = \dfrac{d}{dt}\bigg[v=\sqrt{8s+16}\bigg] = \dfrac{4}{\sqrt{8s+16}} \cdot \dfrac{ds}{dt} = \dfrac{4}{\sqrt{8s+16}} \cdot \sqrt{8s+16} = 4$

(b) you have $v$ as a function of $s$ ... do the graph

(c) $a=4 \implies v = 4t+C$. $s(0) = 6 \implies v(0) = 8 \implies v = ?$ ... then graph $v$ as a function of time.

4. ## Re: Implicit Derivatives?

i can use this as v as a function of s ? --> v(s) = sqrt(8s+16)

and then for the v as function of t ? How do you know that it's linear?

5. ## Re: Implicit Derivatives?

Originally Posted by lc99
i can use this as v as a function of s ? --> v(s) = sqrt(8s+16)
yes

and then for the v as function of t ? How do you know that it's linear?
because acceleration is a constant

6. ## Re: Implicit Derivatives?

Oh my goodness, thanks so much. It seems obvious but not to me! It makes sense now. Thanks!