Fundamental Applied Maths Solutions ›

$$ T(t) = 20 + 60e^-kt $$ This shows the temperature starts at 80 and asymptotically approaches 20 over time.

Vectors are the language of physics and engineering. They describe quantities that have both magnitude and direction.

Solve: ( V(t) = V_0\left( 1 - e^-t/RC \right) ).

Boundary term: ( -\frac\pi\cos(n\pi)n - \left( -\frac(-\pi)\cos(-n\pi)n \right) = -\frac\pi\cos(n\pi)n - \frac\pi\cos(n\pi)n ) (since ( \cos(-n\pi)=\cos(n\pi) )) ( = -\frac2\pi\cos(n\pi)n ).

$$ W = \vecF \cdot \vecd $$ $$ W = (3\hati + 4\hatj) \cdot (3\hati + 4\hatj) $$ $$ W = (3 \times 3) + (4 \times 4) $$ $$ W = 9 + 16 = 25 \text Joules $$

$$ T(t) = 20 + 60e^-kt $$ This shows the temperature starts at 80 and asymptotically approaches 20 over time.

Vectors are the language of physics and engineering. They describe quantities that have both magnitude and direction.

Solve: ( V(t) = V_0\left( 1 - e^-t/RC \right) ).

Boundary term: ( -\frac\pi\cos(n\pi)n - \left( -\frac(-\pi)\cos(-n\pi)n \right) = -\frac\pi\cos(n\pi)n - \frac\pi\cos(n\pi)n ) (since ( \cos(-n\pi)=\cos(n\pi) )) ( = -\frac2\pi\cos(n\pi)n ).

$$ W = \vecF \cdot \vecd $$ $$ W = (3\hati + 4\hatj) \cdot (3\hati + 4\hatj) $$ $$ W = (3 \times 3) + (4 \times 4) $$ $$ W = 9 + 16 = 25 \text Joules $$