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Calculus 1 introduces limits, derivatives, and the basics of integration. Exams test both conceptual understanding — epsilon-delta definitions, the Intermediate Value Theorem — and computational fluency with differentiation rules and curve analysis. Regular timed practice is the fastest way to close gaps before the final.

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Topics typically tested

  • Limits & continuity
  • Differentiation rules (chain, product, quotient)
  • Implicit differentiation
  • Related rates
  • Curve sketching & optimization
  • Fundamental Theorem of Calculus

Sample Calculus 1 exam questions

Three original questions in the style of a real Calculus 1 exam, each with a worked solution. Your own practice exam can be built from your course's notes and syllabus.

Question 1Multiple choiceRelated rates

Water is pumped into an inverted conical tank (vertex at the bottom) at a constant rate of 2 m³/min. The tank is 6 m tall and its circular top has radius 2 m. How fast is the water level rising at the instant the water is 4 m deep?

  1. A.

    38π\dfrac{3}{8\pi} m/min

  2. B.

    98π\dfrac{9}{8\pi} m/min

  3. C.

    12π\dfrac{1}{2\pi} m/min

  4. D.

    172π\dfrac{1}{72\pi} m/min

Worked solution
Answer:B

Let hh be the water depth and rr the radius of the water surface at time tt. Similar triangles give rh=26\dfrac{r}{h} = \dfrac{2}{6}, so r=h3r = \dfrac{h}{3}; this lets the volume be written in terms of hh alone, which is the key step in any related-rates problem.

V=13πr2h=13π(h3)2h=πh327.V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{h}{3}\right)^2 h = \frac{\pi h^3}{27}.

Differentiate with respect to tt (chain rule, because hh depends on tt):

dVdt=πh29 dhdt.\frac{dV}{dt} = \frac{\pi h^2}{9}\,\frac{dh}{dt}.

Substitute the known rate dVdt=2\dfrac{dV}{dt} = 2 m³/min and the instant h=4h = 4 m:

2=π(4)29 dhdt=16π9 dhdt⟹dhdt=1816π=98π≈0.358 m/min.2 = \frac{\pi (4)^2}{9}\,\frac{dh}{dt} = \frac{16\pi}{9}\,\frac{dh}{dt} \quad\Longrightarrow\quad \frac{dh}{dt} = \frac{18}{16\pi} = \frac{9}{8\pi} \approx 0.358\ \text{m/min}.

The most tempting wrong answer is 12π\dfrac{1}{2\pi}, which comes from treating the tank as a cylinder of radius 2 m (V=4πhV = 4\pi h); but the water surface shrinks as the level drops, so rr must be tied to hh before differentiating. Dropping the 13\tfrac13 in the cone volume gives 38π\dfrac{3}{8\pi}, and inverting the similar-triangle ratio (r=3hr = 3h) gives 172π\dfrac{1}{72\pi}.

The water level is rising at 98π≈0.358\dfrac{9}{8\pi} \approx 0.358 m/min (option B).

Question 2Short answerLimits & continuity

The function

f(x)={x+5−3x−4x≠4cx=4f(x) = \begin{cases} \dfrac{\sqrt{x+5} - 3}{x - 4} & x \neq 4 \\[4pt] c & x = 4 \end{cases}

is continuous at x=4x = 4. Find the value of cc.

Worked solution
Answer:

c=16c = \dfrac{1}{6}

Continuity at x=4x = 4 requires f(4)=lim⁡x→4f(x)f(4) = \displaystyle\lim_{x \to 4} f(x), so cc must equal the limit of the first branch. Substituting x=4x = 4 directly gives 00\dfrac{0}{0}, an indeterminate form, so the expression must be simplified first. Multiply numerator and denominator by the conjugate x+5+3\sqrt{x+5} + 3 to clear the square root from the numerator:

x+5−3x−4⋅x+5+3x+5+3=(x+5)−9(x−4)(x+5+3)=x−4(x−4)(x+5+3).\frac{\sqrt{x+5} - 3}{x - 4}\cdot\frac{\sqrt{x+5} + 3}{\sqrt{x+5} + 3} = \frac{(x + 5) - 9}{(x - 4)\left(\sqrt{x+5} + 3\right)} = \frac{x - 4}{(x - 4)\left(\sqrt{x+5} + 3\right)}.

For x≠4x \neq 4 the factor x−4x - 4 cancels, leaving 1x+5+3\dfrac{1}{\sqrt{x+5} + 3}, which is continuous at 4 and can be evaluated by substitution:

lim⁡x→4f(x)=19+3=16.\lim_{x \to 4} f(x) = \frac{1}{\sqrt{9} + 3} = \frac{1}{6}.

Setting f(4)f(4) equal to this limit makes the function continuous there. (Numerically, f(4.001)≈0.16666f(4.001) \approx 0.16666, in agreement.)

c=16c = \dfrac{1}{6}

Question 3Worked problemCurve sketching & optimization

A closed cylindrical can must hold 500 cm³. The material for the top and bottom costs twice as much per square centimetre as the material for the curved side. Find the radius and height (in cm, to 2 decimal places) that minimize the cost of the can, and show that your answer is a minimum.

Worked solution
Answer:

r=(125π)1/3≈3.41r = \left(\dfrac{125}{\pi}\right)^{1/3} \approx 3.41 cm, h=4r≈13.66h = 4r \approx 13.66 cm

Let the radius be rr and the height hh (cm). Let the side material cost kk per cm², so the top and bottom cost 2k2k per cm². The cost is

C=2k (2πr2)+k (2πrh)=4πkr2+2πkrh.C = 2k\,(2\pi r^2) + k\,(2\pi r h) = 4\pi k r^2 + 2\pi k r h.

The volume constraint πr2h=500\pi r^2 h = 500 gives h=500πr2h = \dfrac{500}{\pi r^2}; substituting removes hh so that CC depends on one variable:

C(r)=4πkr2+2πkr⋅500πr2=k(4πr2+1000r),r>0.C(r) = 4\pi k r^2 + 2\pi k r\cdot\frac{500}{\pi r^2} = k\left(4\pi r^2 + \frac{1000}{r}\right), \qquad r > 0.

Differentiate and set the derivative to zero to find critical points:

C′(r)=k(8πr−1000r2)=0⟹r3=10008π=125π⟹r=(125π)1/3=5π1/3≈3.41 cm.C'(r) = k\left(8\pi r - \frac{1000}{r^2}\right) = 0 \quad\Longrightarrow\quad r^3 = \frac{1000}{8\pi} = \frac{125}{\pi} \quad\Longrightarrow\quad r = \left(\frac{125}{\pi}\right)^{1/3} = \frac{5}{\pi^{1/3}} \approx 3.41\ \text{cm}.

The second derivative C′′(r)=k(8π+2000r3)C''(r) = k\left(8\pi + \dfrac{2000}{r^3}\right) is positive for every r>0r > 0, so CC is concave up on its whole domain and this single critical point is the absolute minimum (also, C→∞C \to \infty as r→0+r \to 0^+ and as r→∞r \to \infty).

The height follows from the constraint:

h=500πr2=500π⋅π2/325=20π1/3=4r≈13.66 cm.h = \frac{500}{\pi r^2} = \frac{500}{\pi}\cdot\frac{\pi^{2/3}}{25} = \frac{20}{\pi^{1/3}} = 4r \approx 13.66\ \text{cm}.

So the cheapest can is four times as tall as its radius (twice as tall as it is wide), which makes sense: the expensive end discs are kept small.

r=(125π)1/3≈3.41r = \left(\dfrac{125}{\pi}\right)^{1/3} \approx 3.41 cm and h=4r≈13.66h = 4r \approx 13.66 cm.

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