Question

Can someone help me with a calculus test practice/review?

Answer 1

@phi @ParthKohli @zepdrix

Answer 2

@Luigi0210

Answer 3

@zepdrix that's ok.... want to do a twidla i have 12 questions XD

Answer 4

All you need to do is:
1) Integrate both sides with respect to z.
2) Use the fact that G(1)=8, to solve for the integration constant C.

Answer 5

For example,

\(\color{#000000 }{ \displaystyle \frac{dG}{dz}=4z^2;\quad G(0)=\pi }\)

\(\color{#000000 }{ \displaystyle \int \frac{dG}{dz}~dz=\int 4z^2~dz }\)

Read Carefully. The integral of the derivative of the function, is the function (itself). E.g. \(\color{#000000 }{ \int f'(x)~dx=f(x)\color{grey}{+C} }\).

Answer 6

And then, in the same example you get:
\(\color{#000000 }{ \displaystyle G(z)=\frac{4}{3}z^3+C }\)

Now, use the initial value:
\(\color{#000000 }{ \displaystyle G(0)=\frac{4}{3}0^3+C=\pi \quad \Longrightarrow \quad C=\pi }\)

So, the solution is:
\(\color{#000000 }{ \displaystyle G(z)=\frac{4}{3}z^3+\pi }\)