Question
Ohm’s law states that if the resistance of the path of current between two points is constant, then the voltage difference V between the points and the current I flowing between the points, measured in amperes, satisfy the relationship V=cI, where c is a constant. Which of the following best describes the relationship between the rate of change with respect to time t of the voltage and the rate of change with respect to time t of the current?
A \(\frac{\mathrm{d} V}{\mathrm{d} t}=I\frac{\mathrm{d} c}{\mathrm{d} t}\)
B \(\frac{\mathrm{d} V}{\mathrm{d} t}=c\frac{\mathrm{d} I}{\mathrm{d} t}\)
C \(\frac{\mathrm{d} V}{\mathrm{d} I}=c\)
D \(1=c\frac{\mathrm{d} I}{\mathrm{d} V}\)
▶️ Answer/Explanation
Question
A triangle has base b centimeters and height h centimeters, where the height is three times the base. Both b and h are functions of time t, measured in seconds. If A represents the area of the triangle, which of the following gives the rate of change of A with respect to t ?
A \(\frac{\mathrm{d} A}{\mathrm{d} t}\)=3bcm/sec
B \(\frac{\mathrm{d} A}{\mathrm{d} t}=2b \frac{\mathrm{d} b}{\mathrm{d} t}cm^{2}/sec\)
C \(\frac{\mathrm{d} A}{\mathrm{d} t}=3b \frac{\mathrm{d} b}{\mathrm{d} t}cm/sec\)
D \(\frac{\mathrm{d} A}{\mathrm{d} t}=3b \frac{\mathrm{d} b}{\mathrm{d} t}cm^{2}/sec\)
▶️ Answer/Explanation
Step 1: Express area in terms of base
Given that height h = 3b, the area A is:
\[ A = \frac{1}{2}bh = \frac{1}{2}b(3b) = \frac{3}{2}b^2 \]Step 2: Differentiate area with respect to time
Using the chain rule:
\[ \frac{\mathrm{d} A}{\mathrm{d} t} = \frac{3}{2} \cdot 2b \cdot \frac{\mathrm{d} b}{\mathrm{d} t} = 3b \frac{\mathrm{d} b}{\mathrm{d} t} \]Step 3: Verify units
The units of \( \frac{\mathrm{d} A}{\mathrm{d} t} \) should be:
\[ \text{cm} \times \text{cm/sec} = \text{cm}^2/\text{sec} \]This matches option D.
Conclusion:
The correct expression for the rate of change of the area is:
\[ \frac{\mathrm{d} A}{\mathrm{d} t} = 3b \frac{\mathrm{d} b}{\mathrm{d} t} \text{ cm}^2/\text{sec} \]✅ Answer: D) \( \frac{\mathrm{d} A}{\mathrm{d} t} = 3b \frac{\mathrm{d} b}{\mathrm{d} t} \text{ cm}^2/\text{sec} \)
Question
At a concert, a band is playing on a platform that extends P feet from a wall behind the band, and the platform is rising from ground level, as shown in the figure above. A light source is L feet from the wall, and the platform casts a lengthening shadow on the wall as the platform rises. At time t seconds, the platform is h feet above the ground, and the height of the shadow is s feet. The quantities are related by the equation \[\frac{1}{L}(h+s)=\frac{1}{P}s\] where L and P are constants. Which of the following best expresses the rate of change of h with respect to time in terms of the rate of change of s with respect to time?
A \(\frac{\mathrm{d} h}{\mathrm{d} t}=\frac{L}{P}s-s\)
B \(\frac{\mathrm{d} h}{\mathrm{d} t}=\frac{L}{P}s-\frac{\mathrm{d} s}{\mathrm{d} t}\)
C \(\frac{\mathrm{d} h}{\mathrm{d} t}=\frac{L}{P}\frac{\mathrm{d} s}{\mathrm{d} t}-s\)
D \(\frac{\mathrm{d} h}{\mathrm{d} t}=\frac{L}{P}\frac{\mathrm{d} s}{\mathrm{d} t}-\frac{\mathrm{d} s}{\mathrm{d} t}\)
▶️Answer/Explanation
Ans:D
Differentiating both sides of the given equation with respect to time yields \[\frac{1}{L}(\frac{\mathrm{d} h}{\mathrm{d} t}+\frac{\mathrm{d} s}{\mathrm{d} t})=\frac{1}{P}\frac{\mathrm{d} s}{\mathrm{d} t}\]. Solving for \(\frac{\mathrm{d} h}{\mathrm{d} t}\) gives \(\frac{\mathrm{d} h}{\mathrm{d} t}=\frac{L}{P}\frac{\mathrm{d} s}{\mathrm{d} t}-\frac{\mathrm{d} s}{\mathrm{d} t}\).