Calculus: Derivatives and Kinematics

At $t=3$ hours, the temperature is decreasing at $4$ degrees Celsius per hour. The derivative represents the instantaneous rate of temperature change, not the temperature itself.

At $t=3$ seconds, the particle has acceleration $2\ \mathrm{m/s^2}$ in the positive direction. Since $a(t)=s''(t)$, this describes the instantaneous rate of change of velocity.

$f(a)$ gives the quantity at $x=a$. $f'(a)$ gives its instantaneous rate of change. $f''(a)$ gives the instantaneous rate of change of that rate. Their units are output units, output units per input unit, and output units per squared input unit.

$\displaystyle \frac{\mathrm{cm^3}}{\mathrm{s}}$. The derivative represents cubic centimeters of volume change per second.

$\displaystyle \frac{\mathrm{cm^3}}{\mathrm{s^2}}$. The second derivative represents the rate of change of the volume-change rate.

Original: $s(t)=t^3-6t^2+5$. Differentiate position: $\displaystyle v(t)=\frac{ds}{dt}$. $\displaystyle v(t)=3t^2-12t$. $\boxed{v(t)=3t^2-12t}$. Final Answer: $\boxed{3t^2-12t}$.

Original: $s(t)=t^3-6t^2+5$. First derivative: $v(t)=3t^2-12t$. Differentiate again: $a(t)=6t-12$. $\boxed{a(t)=6t-12}$. Final Answer: $\boxed{6t-12}$.

Original: $s(t)=t^3-6t^2+5$. $v(t)=3t^2-12t$. Set $v(t)=0$: $3t(t-4)=0$. Therefore $t=0$ or $t=4$. $\boxed{t=0,\ 4}$. Final Answer: $\boxed{0,\ 4}$.

Original: $v(t)=3t(t-4)$. For $0<t<4$, $v(t)<0$, so motion is in the negative direction. For $t>4$, $v(t)>0$, so motion is in the positive direction. Final Answer: Negative direction on $(0,4)$ and positive direction on $(4,\infty)$.

No. A particle changes direction only if velocity changes sign. For example, $v(t)=t^2$ gives $v(0)=0$, but $v(t)>0$ on both sides of $t=0$. Therefore the particle does not change direction.

Original: $v=-2$, $a=-3$. Evaluate the sign product: $va=(-2)(-3)=6>0$. Velocity and acceleration have the same sign, so speed is increasing. Final Answer: $\boxed{\text{Speeding up}}$.

Original: $v=-2$, $a=3$. Evaluate the sign product: $va=(-2)(3)=-6<0$. Velocity and acceleration have opposite signs, so speed is decreasing. Final Answer: $\boxed{\text{Slowing down}}$.

No. Positive acceleration means velocity is increasing, not necessarily speed. If $v<0$ and $a>0$, velocity moves toward zero and the particle slows down. If $v>0$ and $a>0$, the particle speeds up.

No. Velocity describes the direction of motion, while acceleration describes how velocity changes. A particle can move in the negative direction while accelerating positively, negatively, or with zero acceleration.

Original: $x^2+y^2=100$. Differentiate both sides with respect to $t$: $\displaystyle \frac{d}{dt}(x^2)+\frac{d}{dt}(y^2)=\frac{d}{dt}(100)$. Apply the chain rule: $\displaystyle 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. Final Answer: $\boxed{2x\frac{dx}{dt}+2y\frac{dy}{dt}=0}$.

Original: $x^2+y^2=100$. Differentiate: $\displaystyle 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. Rearrange: $\displaystyle 2y\frac{dy}{dt}=-2x\frac{dx}{dt}$. Divide by $2y$: $\displaystyle \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$, where $y\neq0$. Final Answer: $\boxed{\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}}$.

Because substituting an instantaneous numerical value before differentiation may incorrectly turn a changing quantity into a constant. First establish the relationship, differentiate with respect to time, and then substitute values for the specified instant.

Original: $A=\pi r^2$. Differentiate with respect to time: $\displaystyle \frac{dA}{dt}=\pi(2r)\frac{dr}{dt}$. Simplify: $\displaystyle \frac{dA}{dt}=2\pi r\frac{dr}{dt}$. Final Answer: $\boxed{2\pi r\frac{dr}{dt}}$.

Original: $\displaystyle V=\frac{4}{3}\pi r^3$. Differentiate: $\displaystyle \frac{dV}{dt}=\frac{4}{3}\pi(3r^2)\frac{dr}{dt}$. Simplify: $\displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$. Final Answer: $\boxed{4\pi r^2\frac{dr}{dt}}$.

Original: $V=\pi r^2h$. Apply the product and chain rules: $\displaystyle \frac{dV}{dt}=\pi\left(2rh\frac{dr}{dt}+r^2\frac{dh}{dt}\right)$. Simplify: $\displaystyle \frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2\frac{dh}{dt}$. Final Answer: $\boxed{2\pi rh\frac{dr}{dt}+\pi r^2\frac{dh}{dt}}$.

Original: $A=LW$. Apply the product rule: $\displaystyle \frac{dA}{dt}=\frac{dL}{dt}W+L\frac{dW}{dt}$. Final Answer: $\boxed{W\frac{dL}{dt}+L\frac{dW}{dt}}$.

Original: $x^2+y^2=L^2$. Since $L$ is constant, $\frac{dL}{dt}=0$. Differentiate: $\displaystyle 2x\frac{dx}{dt}+2y\frac{dy}{dt}=2L(0)$. Final Answer: $\boxed{2x\frac{dx}{dt}+2y\frac{dy}{dt}=0}$.

If $h$ is water height measured upward and the water level is falling, then $\frac{dh}{dt}<0$. The negative sign indicates decreasing height. A complete AP response must include the rate's magnitude and units.

If $\frac{dV}{dt}>0$, volume is increasing at that instant. For example, $\frac{dV}{dt}=8\ \mathrm{cm^3/s}$ means volume increases at $8$ cubic centimeters per second.

If $\frac{dV}{dt}<0$, volume is decreasing at that instant. For example, $\frac{dV}{dt}=-8\ \mathrm{cm^3/s}$ means volume decreases at $8$ cubic centimeters per second.

Units distinguish quantities from their rates. A radius might be measured in $\mathrm{cm}$, while $\frac{dr}{dt}$ is measured in $\mathrm{cm/s}$. A volume-change rate is measured in $\mathrm{cm^3/s}$, not $\mathrm{cm^3}$.

Original: $y=f(x)$. The tangent line at $x=a$ has slope $f'(a)$ and passes through $(a,f(a))$. Point-slope form: $L(x)-f(a)=f'(a)(x-a)$. Rearrange: $\boxed{L(x)=f(a)+f'(a)(x-a)}$. Final Answer: $\boxed{f(a)+f'(a)(x-a)}$.

Original: $L(x)=f(a)+f'(a)(x-a)$. Substitute $x=a+h$: $L(a+h)=f(a)+f'(a)((a+h)-a)$. Simplify: $L(a+h)=f(a)+f'(a)h$. Final Answer: $\boxed{f(a+h)\approx f(a)+f'(a)h}$.

Original: $f(x)=\sqrt{x}$, $a=4$. $f(4)=2$. $\displaystyle f'(x)=\frac{1}{2\sqrt{x}}$. $\displaystyle f'(4)=\frac14$. Substitute into $L(x)=f(a)+f'(a)(x-a)$: $\displaystyle L(x)=2+\frac14(x-4)$. Final Answer: $\boxed{L(x)=2+\frac{x-4}{4}}$.

Original: $f(x)=\sqrt{x}$, $a=4$. $\displaystyle L(x)=2+\frac{x-4}{4}$. Substitute $x=4.1$: $\displaystyle L(4.1)=2+\frac{4.1-4}{4}$. $=2+\frac{0.1}{4}$. $=2+0.025$. $=2.025$. Final Answer: $\boxed{\sqrt{4.1}\approx2.025}$.

A differentiable function and its linearization have the same value and first derivative at the tangency point. Formally, $\displaystyle \lim_{h\to0}\frac{f(a+h)-L(a+h)}{h}=0$. The error becomes small relative to the input change.

Original: $y=f(x)$. Differentiate to obtain $f'(x)$. Multiply by the independent differential $dx$. Final Answer: $\boxed{dy=f'(x)\,dx}$.

The actual change is $\displaystyle \Delta y=f(x+\Delta x)-f(x)$. The differential estimate is $\displaystyle dy=f'(x)\Delta x$. For small $\Delta x$, $\Delta y\approx dy$, but they are not generally equal.

Original: $y=x^3$, $x=2$, $dx=0.01$. Differentiate: $\displaystyle dy=3x^2dx$. Substitute: $dy=3(2)^2(0.01)$. $=3(4)(0.01)$. $=0.12$. Final Answer: $\boxed{dy=0.12}$.

If $f''(x)>0$ throughout the relevant interval, the function is concave up and the tangent line underestimates. If $f''(x)<0$ throughout the interval, the function is concave down and the tangent line overestimates.

When $f''(x)>0$ throughout the relevant interval, the derivative increases and the graph lies above its tangent line. Therefore $f(x)\geq L(x)$, so $L(x)$ underestimates or equals the actual value.

When direct limit analysis produces $\frac00$ or $\frac{\infty}{\infty}$ and the rule's differentiability, nonzero denominator-derivative, and derivative-quotient limit conditions hold. Then $\displaystyle \lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}$.

Original: $\displaystyle \lim_{x\to0}\frac{\sin x}{x}$. Direct substitution gives $\frac00$. Differentiate numerator and denominator separately: $\displaystyle \lim_{x\to0}\frac{\cos x}{1}$. Substitute $x=0$: $\cos0=1$. Final Answer: $\boxed{1}$.

Original: $\displaystyle \lim_{x\to0}\frac{e^x-1}{x}$. Direct substitution gives $\frac00$. Differentiate numerator and denominator: $\displaystyle \lim_{x\to0}\frac{e^x}{1}$. Substitute $x=0$: $e^0=1$. Final Answer: $\boxed{1}$.

Original: $\displaystyle \lim_{x\to\infty}\frac{x}{e^x}$. The form is $\frac{\infty}{\infty}$. Differentiate numerator and denominator: $\displaystyle \lim_{x\to\infty}\frac{1}{e^x}$. Since $e^x\to\infty$, the quotient approaches zero. Final Answer: $\boxed{0}$.

The rule requires an applicable indeterminate form such as $\frac00$ or $\frac{\infty}{\infty}$. If direct substitution gives a finite quotient with a nonzero denominator, evaluate it directly rather than applying L'Hôpital's Rule.

No. A product with form $0\cdot\infty$ must first be rewritten as a quotient. For example, $f(x)g(x)=\frac{f(x)}{1/g(x)}$. If the transformed quotient has an applicable indeterminate form and the rule's conditions hold, L'Hôpital's Rule may be used.

No. First combine fractions, rationalize, factor, or otherwise transform the expression into an applicable quotient. Then check whether the result has form $\frac00$ or $\frac{\infty}{\infty}$.

After each application, evaluate the new quotient's limiting form. If it remains $\frac00$ or $\frac{\infty}{\infty}$ and the rule's hypotheses hold, apply the rule again. Stop when the limit can be evaluated or another method is required.

The quotient rule differentiates a quotient: $\displaystyle \left(\frac fg\right)'=\frac{gf'-fg'}{g^2}$. L'Hôpital's Rule evaluates certain indeterminate limits using $\displaystyle \lim\frac fg=\lim\frac{f'}{g'}$ under its hypotheses. These are different operations.

Because a derivative generally varies with the input. For example, $V'(5)=12$ means the instantaneous volume-change rate at $t=5$ is $12$ volume units per time unit, not that the rate is always $12$.

Find times where $v(t)=0$ or velocity is undefined. Test the sign of $v(t)$ on both sides of each candidate. A change from positive to negative or negative to positive indicates a direction change.

For $v(t)\neq0$, calculate or compare the signs of velocity and acceleration. If $v(t)a(t)>0$, the particle speeds up. If $v(t)a(t)<0$, it slows down.

Acceleration is the slope of the velocity graph: $\displaystyle a(t)=v'(t)$. Positive slope means positive acceleration; negative slope means negative acceleration; horizontal tangent means zero acceleration.

Velocity is the slope of the position graph: $\displaystyle v(t)=s'(t)$. Positive slope means motion in the positive direction; negative slope means motion in the negative direction; horizontal tangent means instantaneous rest.

Define changing variables and units; write the governing equation; differentiate with respect to time before substituting instantaneous values; substitute known quantities and rates; solve algebraically; include correct units; and interpret the sign and physical meaning.

State the input, the changing quantity, the numerical rate, the units, and whether the quantity increases or decreases. Example: If $V'(5)=-12\ \mathrm{L/min}$, then at $t=5$ minutes, volume is decreasing at $12$ liters per minute.