Question

I have
`f(t)= ( t^(2) )/(9)` and
`g(t)= e^{ (- ln(t))/(4) }`

Given that
`k(t)=f(t)+cg(t)` and k(1)=4, what is the value of c?

Answer 1

`\bf \begin{cases} f(t)=\cfrac{t^2}{9}\\\\ g(t)=e^{\cfrac{}{}(-ln(t))/(4)}\\\\ k(t)=f(t)+c\cdot g(t) \end{cases}\implies k(t)=\cfrac{t^2}{9}+C\cdot e^{\cfrac{}{}(-ln(t))/(4)} \\\\\\ k(1)=4\implies \begin{cases} k=4\\ t=1 \end{cases}\implies 4=\cfrac{1^2}{9}+C\cdot e^{\cfrac{}{}(-ln(1))/(4)} \\\\\\ 4=\cfrac{1}{9}+C\cdot e^{(0)/(4)}\implies 4=\cfrac{1}{9}+C\cdot 1\implies 4-\cfrac{1}{9}=C \\\\\\ \cfrac{35}{9}=C`