AP Calculus BC

AP Calculus BC Practice Test: Practice Test 6

Practice Test 6 for AP Calculus BC: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

Which integral represents area between $x=10-y$ (right) and $x=2y^2$ (left) on $0\le y\le 2$ ?

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Question 1

Which integral represents area between $x=10-y$ (right) and $x=2y^2$ (left) on $0\le y\le 2$ ?

$\displaystyle \int_{0}^{2}\big[2y^2-(10-y)\big]dy$
$\displaystyle \int_{0}^{2}\big[(10-y)-2y^2\big]dy$ (correct answer)
$\displaystyle \int_{2}^{0}\big[(10-y)-2y^2\big]dy$
$\displaystyle \int_{0}^{2}\big[(10-y)+2y^2\big]dy$
$\displaystyle \int_{-2}^{0}\big[(10-y)-2y^2\big]dy$

Explanation: This problem practices dy area setups. Integral of right minus left. Right: x = 10 - y, left: x = 2y², so (10 - y) - 2y². From 0 to 2. Choice A subtracts wrong, negative. Compare at y=1.

Question 2

A particle moves with $r(t)=\langle t^2-1,\,3t\rangle$ for $0\le t\le2$ ; what is its speed at $t=1$ ?

$\sqrt{13}$ (correct answer)
$\sqrt{10}$
$13$
$10$
$3$

Explanation: This problem involves analyzing the motion of a particle using parametric equations. To find the speed, first compute the velocity vector by differentiating each component of the position vector $r(t) = \langle t^2 - 1, 3t \rangle$ with respect to t, yielding $v(t) = \langle 2t, 3 \rangle$ . Then, evaluate the velocity at t = 1 to get $v(1) = \langle 2, 3 \rangle$ . The speed is the magnitude of this velocity vector, calculated as $\sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$ . A tempting distractor is 13, which arises from adding the squared components without taking the square root, thus confusing speed with the square of the magnitude. A transferable motion-analysis strategy is to always differentiate position to get velocity and then compute its magnitude for speed in parametric equations.

Question 3

If $f(x)=\sqrt{7-3x}$ gives concentration, what is $f'(x)$ ?

$\dfrac{1}{2\sqrt{7-3x}}$
$\dfrac{3}{2\sqrt{7-3x}}$
$-\dfrac{3}{2\sqrt{7-3x}}$ (correct answer)
$-\dfrac{1}{2\sqrt{7-3x}}$
$-\dfrac{3}{\sqrt{7-3x}}$

Explanation: This problem requires the chain rule to differentiate a square root function. The function f(x) = √(7 - 3x) = (7 - 3x)^(1/2) has an outer function g(u) = u^(1/2) and an inner function u = 7 - 3x. By the chain rule, f'(x) = (1/2)u^(-1/2) · u'(x) = (1/2)(7 - 3x)^(-1/2) · (-3) = -3/(2√(7 - 3x)). Option B shows 3/(2√(7 - 3x)), missing the negative sign from the derivative of (7 - 3x). Always pay attention to the sign when differentiating expressions with subtraction, as the chain rule preserves these signs through multiplication.

Question 4

A decay model uses $h(x)=e^{-x}$ for $x\ge0$ ; does $\int_{0}^{\infty} h(x)\,dx$ converge, and to what value?

Converges to $1$ (correct answer)
Converges to $0$
Diverges
Converges to $e$
Converges to $\tfrac{1}{e}$

Explanation: Evaluating improper integrals involves determining whether they converge and finding their value if they do, often by taking limits. To assess ∫ from 0 to ∞ of e^{-x} dx, express it as the limit as b approaches infinity of the integral from 0 to b. The antiderivative is -e^{-x}, so evaluate lim_{b→∞} [-e^{-b} + e^{0}] which is 0 + 1 = 1. Thus, the integral converges to 1. A tempting distractor might be to think it converges to 0 because e^{-∞}=0, but forgetting the +1 from the lower bound leads to that error. A general strategy for improper integrals with exponential decay is to recognize their rapid convergence and compute limits directly.

Question 5

Let $f(x)=x^3$ and $g(x)=x^2$ . Which setup gives total area between the curves on $[0,1]$ ?

$\displaystyle \int_{0}^{1}(x^3-x^2)\,dx$
$\displaystyle \int_{0}^{1}(x^2-x^3)\,dx$ (correct answer)
$\displaystyle \int_{0}^{1}(x^3+x^2)\,dx$
$\displaystyle \int_{0}^{0}(x^2-x^3)\,dx+\int_{0}^{1}(x^3-x^2)\,dx$
$\displaystyle \int_{0}^{1}(x^3-1)\,dx$

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curves f(x) = x³ and g(x) = x² intersect at x = 0 and 1, with g(x) above f(x) throughout (0, 1), so no internal crossing requires splitting beyond the single interval [0, 1]. Since they touch at endpoints and g > f inside, we integrate g - f = x² - x³ over [0, 1]. No further splitting is needed as the top function remains consistent. A tempting distractor like choice A fails because it integrates f - g, which is negative throughout, giving a negative value instead of positive total area. To find areas between intersecting curves generally, always locate intersection points, test which function is greater in each subinterval, and integrate the positive difference in each.

Question 6

Given $L(x)=\dfrac{x^3-5}{2x^4+1}$ , what is the horizontal asymptote as $x\to\infty$ ?

$y=0$ (correct answer)
$y=\dfrac{1}{2}$
$y=\infty$
$x=0$
$x=-\sqrt[4]{\dfrac{1}{2}}$

Explanation: This problem involves finding limits at infinity to determine horizontal asymptotes. For L(x) = (x³ - 5)/(2x⁴ + 1), numerator degree 3 less than denominator 4, so horizontal asymptote y = 0 as x approaches infinity. Denominator dominates. Same for both directions. A tempting distractor is y = 1/2, misusing leading coefficients. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if the numerator's degree is lower, the asymptote is y = 0.

Question 7

Compute $\displaystyle\lim_{x\to 0}\dfrac{\sin(2x)-2\sin x}{x^3}$ for a third-order comparison near zero.

$0$
$1$
$-1$ (correct answer)
$2$
$-2$

Explanation: This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form. As x approaches 0, sin(2x) - 2 sin(x) approaches 0 - 0 = 0 and x^3 approaches 0, yielding 0/0. Derivatives: 2 cos(2x) - 2 cos(x) over 3x^2, still 0/0 (21 - 21=0); again: -4 sin(2x) + 2 sin(x) over 6x, still 0/0; again: -8 cos(2x) + 2 cos(x) over 6 = (-81 + 21)/6 = -6/6 = -1. Repeated application is needed due to persistent indeterminacy. A tempting distractor is 0, from direct substitution without accounting for higher derivatives. To recognize indeterminate forms like 0/0, always plug in the limit value to check if both numerator and denominator approach 0 or infinity.

Question 8

Let $u(x)=\begin{cases}\ln(x),&x>0\\0,&x\le0\end{cases}$ . What type of discontinuity occurs at $x=0$ ?

Removable discontinuity
Infinite discontinuity (correct answer)
Oscillating discontinuity
Jump discontinuity
No discontinuity

Explanation: This problem involves analyzing the piecewise function u(x) where u(x) = ln(x) for x > 0 and u(x) = 0 for x ≤ 0. As x approaches 0 from the right, ln(x) approaches -∞ because the natural logarithm of positive numbers approaching zero becomes increasingly negative without bound. From the left, u(x) = 0 for all x ≤ 0, so lim(x→0⁻) u(x) = 0. Since the right-hand limit approaches -∞ while the left-hand limit is finite, this creates an infinite discontinuity at x = 0. A jump discontinuity would require both one-sided limits to be finite. When analyzing logarithmic functions near their domain boundaries, remember that ln(x) → -∞ as x → 0⁺.

Question 9

Which integral gives the volume when the region between $y=e^x$ and $y=1$ on $[0,\ln 3]$ is revolved about the $x$ -axis?

$\pi\int_{0}^{\ln 3}\big[(1)^2-(e^x)^2\big]dx$
$\pi\int_{0}^{\ln 3}\big[(e^x)^2-(1)^2\big]dx$ (correct answer)
$\pi\int_{0}^{3}\big[(e^y)^2-(1)^2\big]dy$
$\pi\int_{0}^{\ln 3}\big[e^x-1\big]^2dx$
$\pi\int_{0}^{\ln 3}\big[(e^x)-(1)\big]dx$

Explanation: This problem requires the washer method for revolution around the x-axis. The washer method uses π∫[R(x)]² - [r(x)]² dx, where R(x) is the outer radius and r(x) is the inner radius. On [0, ln 3], we need to determine which function is farther from the x-axis. Since e⁰ = 1 and e^(ln 3) = 3, the exponential function y = e^x starts at 1 and increases to 3, while y = 1 is constant. Therefore, e^x ≥ 1 throughout the interval, making e^x the outer radius and 1 the inner radius. The correct integral is π∫₀^(ln 3)[(e^x)² - (1)²]dx. Choice D incorrectly uses (e^x - 1)², squaring the difference instead of taking the difference of squares. When using washers, always square each function separately—this represents the areas of the outer and inner circles whose difference gives the washer area.

Question 10

The height of a plant, $H(t)$ , in centimeters, is a differentiable function of time $t$ in days. Measurements are recorded as follows: $H(10) = 20$ , $H(12) = 24$ , and $H(15)=33$ .

Which of the following is the best estimate of the growth rate $H'(12)$ in centimeters per day?

$2.0$
$2.5$ (correct answer)
$2.6$
$3.0$

Explanation: To get the best estimate for $H'(12)$ , we should average the rates of change from the intervals on either side of $t=12$ . The backward difference (rate on $[10, 12]$ ) is $\frac{H(12)-H(10)}{12-10} = \frac{24-20}{2} = 2$ . The forward difference (rate on $[12, 15]$ ) is $\frac{H(15)-H(12)}{15-12} = \frac{33-24}{3} = 3$ . The best estimate is the average of these two rates: $\frac{2+3}{2} = 2.5$ cm/day.

Question 11

A velocity integral is $\int \frac{\sec^2 x}{\tan x}\,dx$ ; which technique is most appropriate?

Integration by parts
Substitution using $u=\tan x$ (correct answer)
Trigonometric substitution
Partial fractions
Use a series expansion

Explanation: Selecting an appropriate integration technique requires identifying substitution opportunities with trigonometric functions. The integrand $\frac{\sec^2 x}{\tan x}$ can be rewritten as $\frac{\sec^2 x}{\tan x}$ , where the numerator $\sec^2 x$ is the derivative of $\tan x$ . Using substitution $u = \tan x$ gives $du = \sec^2 x\,dx$ , transforming the integral to $\int \frac{1}{u}\,du = \ln|u| + C$ . Trigonometric substitution isn't needed when a simpler algebraic substitution works. When the numerator is the derivative of the denominator (or a constant multiple), substitution using the denominator expression is the most efficient approach.

Question 12

For $s(x)=\frac{x^3-8}{x-2}$ when $x\ne2$ , which expression represents $\lim_{x\to2}s(x)$ ?

$\lim_{x\to2} s(x)=12$ (correct answer)
$s(2)=12$
$\lim_{x\to2} s(x)=6$
$\lim_{x=2} s(x)=12$
$\lim_{x\to12} s(x)=2$

Explanation: This question tests the skill of defining limits and using limit notation. The expression \lim_{x\to2} s(x)=12 is correct because simplifying s(x) to x^2 + 2x + 4 for x≠2 yields 12 at x=2, indicating the limit. As x nears 2, values approach 12 despite the hole. This notation denotes the approaching value accurately. A tempting distractor is B, s(2)=12, but s(2) is undefined, confusing limit with function value. To check limit notation, use → for approach, simplify expressions where possible, and verify the limit value algebraically.

Question 13

In a physics model, accumulated drift is $\sum_{n=1}^{\infty}\frac{1}{(n+1)\sqrt{n+1}}$ . Use the integral test for convergence.

Diverges because $\int_1^{\infty}\frac{1}{(x+1)\sqrt{x+1}}\,dx$ diverges.
Converges because $\int_1^{\infty}\frac{1}{(x+1)\sqrt{x+1}}\,dx$ converges. (correct answer)
Converges because $\int_1^{\infty}\frac{1}{(x+1)\sqrt{x+1}}\,dx=\left.\ln(x+1)\right|_1^{\infty}$ is finite.
Diverges because $\int_1^{\infty}\frac{1}{(x+1)\sqrt{x+1}}\,dx=\left.-\frac{1}{\sqrt{x+1}}\right|_1^{\infty}$ is infinite.
Converges because $\int_1^{\infty}\frac{1}{(x+1)\sqrt{x+1}}\,dx$ is improper at $x=1$ .

Explanation: This problem applies the integral test to determine series convergence. The integrand $\frac{1}{(x+1)\sqrt{x+1}} = \frac{1}{(x+1)^{3/2}}$ suggests the substitution $u = x+1$ . Then $\int_1^{\infty}\frac{1}{(x+1)^{3/2}}\,dx = \int_2^{\infty}\frac{1}{u^{3/2}}\,du = \left[-\frac{2}{\sqrt{u}}\right]_2^{\infty} = 0 - (-\frac{2}{\sqrt{2}}) = \sqrt{2}$ . Since the integral converges to a finite value, the series converges by the integral test. Choice C incorrectly suggests the antiderivative is $\ln(x+1)$ , which would be correct for $\int\frac{1}{x+1}\,dx$ but not for our integral with the additional $\sqrt{x+1}$ factor. When dealing with expressions like $(x+a)^p$ where $p < -1$ , the integral test shows convergence.

Question 14

The cost function is $C(x)=7x^4-5x^2+9$ . What is the third derivative $C^{(3)}(x)$ ?

$168x$ (correct answer)
$84x^2-10$
$168x^2$
$84x$
$168x^2-10$

Explanation: This problem tests your ability to find higher-order derivatives, specifically the third derivative of a cost function. Given C(x) = 7x^4 - 5x^2 + 9, we differentiate step by step: $C'(x) = 28x^3 - 10x$ , $C''(x) = 84x^2 - 10$ , and $C'''(x) = 168x$ . Choice B ( $84x^2 - 10$ ) might seem correct if you stop at the second derivative, a common mistake when tracking multiple differentiations. For polynomial functions, the nth derivative eliminates all terms of degree less than n, which is why the constant and x^2 terms vanish by the third derivative.

Question 15

If $\int_{-3}^{1} f(x)\,dx=2$ and $\int_{-3}^{1} g(x)\,dx=-5$ , what is $\int_{-3}^{1} \big(3f(x)-2g(x)\big)\,dx$ ?

$-4$
$16$ (correct answer)
$4$
$-16$
$11$

Explanation: This problem tests the properties of definite integrals, specifically the linearity property for sums and scalar multiples. The integral of 3f(x) - 2g(x) from -3 to 1 is 3 times the integral of f minus 2 times the integral of g over the same interval. Given integral of f is 2 and g is -5, this computes to $3 \times 2 - 2 \times (-5) = 6 + 10 = 16$ . Thus, the value is 16. A tempting distractor is 4, which might come from mistakenly adding instead of properly applying the coefficients. Remember this transferable checklist for definite integral properties: verify interval additivity, apply linearity for constants and sums, reverse limits with a negative sign, and consider even or odd function symmetries when applicable.

Question 16

For $g(x)=12x^3-5x+7x^2$ , what is $g'(x)$ ?

$36x^2-5+14x$ (correct answer)
$36x^2-5+7x$
$36x^3-5+14x^2$
$12x^2-5+14x$
$36x^2-5+14x^2$

Explanation: Finding g'(x) requires applying the power rule to each term, but first we should rewrite g(x) = 12x^3 - 5x + 7x^2 in standard form. The power rule states that the derivative of x^n is nx^(n-1). Differentiating term by term: 12x^3 becomes 36x^2, 7x^2 becomes 14x, and -5x becomes -5. Choice B incorrectly calculates the derivative of 7x^2 as 7x instead of 14x, forgetting to multiply by the exponent. When applying the power rule, always multiply the coefficient by the current exponent before reducing the exponent by 1.

Question 17

A curve is defined by $x=\tan t$ and $y=\sec t$ ; what is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ?

$\dfrac{\sec t}{\tan t}$
$\dfrac{1}{\sec^3 t}$ (correct answer)
$\dfrac{\sec^2 t}{\tan t}$
$\dfrac{1}{\sec t\tan^2 t}$
$\dfrac{1}{\sec^2 t\tan t}$

Explanation: This problem requires finding the second derivative for trigonometric parametric equations involving tangent and secant. We have $\frac{dx}{dt} = \sec^2 t$ and $\frac{dy}{dt} = \sec t \tan t$ , so $\frac{dy}{dx} = \frac{\sec t \tan t}{\sec^2 t} = \frac{\tan t}{\sec t} = \frac{\sin t}{\cos t} \cdot \cos t = \sin t$ . Differentiating with respect to $t$ gives $\frac{d}{dt}(\sin t) = \cos t$ . Therefore, $\frac{d^2y}{dx^2} = \frac{\cos t}{\sec^2 t} = \cos t \cdot \cos^2 t = \cos^3 t = \frac{1}{\sec^3 t}$ . Choice D might tempt students who make an error with the reciprocal relationships. The key insight is recognizing that $\frac{dy}{dx}$ simplifies to $\sin t$ , making the second derivative calculation straightforward.

Question 18

Use the ratio test to determine convergence of $\sum_{n=1}^{\infty} \frac{n^4}{n!}$ .

Inconclusive because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$ .
Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1$ .
Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1$ .
Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$ . (correct answer)
Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$ .

Explanation: This problem requires applying the ratio test to a series with polynomial numerator and factorial denominator. For $\sum_{n=1}^{\infty} \frac{n^4}{n!}$ , we calculate $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)^4/(n+1)!}{n^4/n!} = \frac{(n+1)^4 \cdot n!}{n^4 \cdot (n+1)!} = \frac{(n+1)^4}{n^4(n+1)} = \frac{(n+1)^3}{n^4} = \frac{1}{n} \cdot \left(\frac{n+1}{n}\right)^3$ . As $n \to \infty$ , this approaches $0 < 1$ , so the series converges. Choice B incorrectly suggests the limit is infinity and divergence, missing that factorial growth dominates polynomial growth. Factorials in denominators almost always lead to convergence.

Question 19

For $x>0$ , minimize $f(x)=\frac{25}{x}+x$ . What value of $x$ minimizes $f$ ?

$25$
$5$ (correct answer)
$1$
$rac{1}{5}$
$0$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For $f(x) = \frac{25}{x} + x$ for $x > 0$ , compute $f'(x) = -\frac{25}{x^2} + 1 = 0$ , giving $x = 5$ . The second derivative $f''(x) = \frac{50}{x^3} > 0$ for $x > 0$ confirms a minimum. As $x$ approaches 0 or infinity, $f$ increases. A tempting distractor is $x = 25$ , but it gives higher value than at $x = 5$ . A transferable optimization-solving strategy is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, evaluate at critical points and endpoints, and use tests to confirm the extremum.

Question 20

A continuous function $u$ on $[-3,1]$ satisfies $u(-3)=0.5$ and $u(1)=2.5$ . Does IVT guarantee a solution to $u(x)=1$ ?

Yes, because $u$ is continuous on $[-3,1]$ and $1$ is between $u(-3)$ and $u(1)$ . (correct answer)
No, because $u(-3)$ is not an integer.
Yes, because $u(-3)<u(1)$ .
No, because IVT requires $u(-3)u(1)<0$ .
Yes, because $u$ is defined for all $x$ between $-3$ and $1$ .

Explanation: This question tests your ability to apply the Intermediate Value Theorem with decimal values. The IVT states that if f is continuous on [a,b] and k is between f(a) and f(b), then f(c) = k for some c in [a,b]. Here, u is continuous on [-3,1] with u(-3) = 0.5 and u(1) = 2.5, and we want u(x) = 1. Since 0.5 < 1 < 2.5, the value 1 is between u(-3) and u(1), so IVT guarantees a solution exists. Choice D incorrectly thinks IVT requires the product u(-3)u(1) < 0, but that condition is only needed when seeking zeros—for general values, we just need the target between endpoints. The IVT checklist: (1) continuous on closed interval, (2) target value between endpoint function values—both conditions are met here.

Question 21

If $s'(1)=0$ and $s''(1)=-4$ , how is the critical point at $x=1$ classified?

Local maximum at $x=1$ (correct answer)
Local minimum at $x=1$
Neither; test inconclusive since $s''(1)<0$
Neither; $s''(1)<0$ implies concave up
Cannot be determined without checking $s'(x)$ near $1$

Explanation: The Second Derivative Test is a method to classify critical points of a function by examining the sign of the second derivative at those points. At x=1, s'(1)=0 marks a critical point, and s''(1)=-4, negative, shows concave down. This suggests a hilltop shape, indicating a local maximum at x=1. The negative value ensures the function decreases on both sides. A tempting distractor is choice D, which claims s''(1)<0 implies concave up, but it actually means concave down. To apply the Second Derivative Test effectively, always ensure the first derivative is zero and check if the second derivative is non-zero; if it is zero, switch to the First Derivative Test for classification.

Question 22

A telescope correction uses $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n^2-1}$ . Does it converge?

Diverges because the series is not defined at $n=1$ . (correct answer)
Converges by the Alternating Series Test because $\frac{1}{n^2-1}$ decreases and approaches $0$ .
Converges absolutely because $\sum \frac{1}{n^2-1}$ converges.
Diverges because $\lim_{n\to\infty}\frac{1}{n^2-1}=1$ .
Diverges because $\frac{1}{n^2-1}$ increases.

Explanation: The skill being tested here is the Alternating Series Test for convergence. The test applies to series of the form $\sum (-1)^{n+1} b_n$ where $b_n > 0$ , $b_n$ decreases monotonically, and $\lim b_n = 0$ . These conditions guarantee the series converges by bounding the partial sums. However, for $\sum (-1)^{n+1} \frac{1}{n^2 - 1}$ , the term at n=1 is undefined due to division by zero, so the series cannot converge. A tempting distractor is choice B, which claims convergence by AST, but it fails because the series isn't even defined for all terms. A transferable strategy for alternating series is to ensure all terms are well-defined before applying any convergence test.

Question 23

Which integral gives the volume if the base is $0\le x\le 3$ , $0\le y\le 2x$ , with right-triangle cross sections perpendicular to $x$ ?

$\displaystyle \int_{0}^{3}\frac{1}{2}(2x)^2\,dx$ (correct answer)
$\displaystyle \int_{0}^{3}\frac{1}{2}(2x)\,dx$
$\displaystyle \int_{0}^{2}\frac{1}{2}\left(\frac{y}{2}\right)^2\,dy$
$\displaystyle \int_{0}^{3}(2x)^2\,dx$
$\displaystyle \int_{0}^{3}\frac{\pi}{8}(2x)^2\,dx$

Explanation: This problem involves finding the volume when cross sections perpendicular to the x-axis are right triangles. The base region has 0 ≤ x ≤ 3 and 0 ≤ y ≤ 2x, so at each x-value, the base width is 2x. For right triangle cross sections with legs equal to the base width, both legs have length 2x. The area of a right triangle is (1/2) × base × height = (1/2)(2x)(2x) = (1/2)(2x)². Choice D incorrectly omits the factor of 1/2 from the triangle area formula. The transferable strategy is that for isosceles right triangles with legs equal to the base width w, the area is (1/2)w².

Question 24

The temperature $T$ , in degrees Celsius, of a chemical reaction is a differentiable function of time $t$ , in minutes. At time $t=5$ minutes, the temperature is $40^{\circ}C$ . At time $t=5.5$ minutes, the temperature is $48^{\circ}C$ .

Based on this information, which of the following is the best estimate for the rate of change of the temperature, in degrees Celsius per minute, at time $t=5$ minutes?

$8$
$16$ (correct answer)
$1.45$
$44$

Explanation: The rate of change of the temperature at $t=5$ can be estimated by the average rate of change over the interval $[5, 5.5]$ . $T'(5) \approx \frac{T(5.5) - T(5)}{5.5 - 5} = \frac{48 - 40}{0.5} = \frac{8}{0.5} = 16$ degrees Celsius per minute.

Question 25

An electrical signal is modeled by $V(t)=\ln\big((t^2+4)^3\big)$ . Which technique is most appropriate to find $V'(t)$ ?

Differentiate using the chain rule with properties of logarithms (correct answer)
Differentiate using the quotient rule
Differentiate using implicit differentiation
Differentiate using the product rule
Differentiate by converting to a Maclaurin series first

Explanation: Selecting the appropriate differentiation technique is a key skill in calculus, ensuring efficiency and accuracy in finding derivatives. For V(t) = ln((t² + 4)^3), using the chain rule with properties of logarithms is most appropriate because simplifying to 3 ln(t² + 4) first allows easy differentiation with the chain rule on the inner function. This approach leverages the logarithm power rule to reduce complexity before applying the derivative. It results in V'(t) = 3 * (1/(t² + 4)) * 2t, which is clean and direct. Differentiating using the product rule would be misguided since the function isn't a product after simplification. Consider logarithmic properties to simplify expressions before differentiation, especially with exponents inside logs, for a more streamlined process.

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