1. ## Calculus Problems

1. if $\displaystyle \frac{dy}{dx}=4y$ and if $\displaystyle y=4$ when $\displaystyle x=0$, then $\displaystyle y=$

2. $\displaystyle \int_0^{1/2} \frac{2x dx}{\sqrt{1-x^2}}=$

3. the point on the curve $\displaystyle 2y=x^2$ nearest to $\displaystyle (4,1)$ is

4. if $\displaystyle tan(xy) = x$, then $\displaystyle \frac{dy}{dx}=$

2. Originally Posted by yamuda
1. if $\displaystyle \frac{dy}{dx}=4y$ and if $\displaystyle y=4$ when $\displaystyle x=0$, then $\displaystyle y=$

2. $\displaystyle \int_0^{1/2} \frac{2x dx}{\sqrt{1-x^2}}=$

3. the point on the curve $\displaystyle 2y=x^2$ nearest to $\displaystyle (4,1)$ is

4. if $\displaystyle tan(xy) = x$, then $\displaystyle \frac{dy}{dx}=$
First one is an SDE so you go $\displaystyle \frac{dy}{y}=4dx$ integrating we get $\displaystyle ln|y|=4x+c$ then you get $\displaystyle y=Ce^{4x}$

2. The derivative of the quantity is present therefore $\displaystyle -\int_0^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}dx=-2\sqrt{1-x^2}|_0^{\frac{1}{2}}$

for the third one I think you can figure that out

4.differenating implicitly we get $\displaystyle sec^2(xy)\cdot{(xy'+y)}=1$ now solve for y'