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Calculus 1 : Calculus

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Example Questions

Example Question #61 : Spatial Calculus

The position of an object is given by the equation x(t) = 4t^2 + 2 . What is the velocity of this object when t = 5 ?

Possible Answers:

Correct answer:

Explanation:

The velocity of the object can be found by differentiating the position equation. This can be done using the power rule where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Appying this to the position equation gives us

$v(t) = 4(2)t^{(2-1)} + 2(1)t^{1-1} = 8t$ .

We can now solve for the velocity to by inputting t = 5 into the velocity equation

v(5) = 8(5) = 40 .

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Example Question #2 : Derivatives

The position of an object is given by the equation $x(t) = 7t^2\sin(t)$ . What is the equation for the velocity of the object?

Possible Answers:

$v(t) = 14t\sin(t) +7t^2\cos(t)$

$v(t) = 14t\sin(t) - 7t^2\cos(t)$

$v(t) = 14t\cos(t)$

$v(t) = 14t\sin^2(t) - 7t\cos(t)$

Correct answer:

$v(t) = 14t\sin(t) +7t^2\cos(t)$

Explanation:

The velocity of the object can be found by differentiating the position equation. The position equation can be accurately differentiated using the power rule and the product rule where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$

and where if

$f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$

Using these two rules we find the velocity equation to be

$v(t) = x'(t) = 7(2)t^{(2-1)}(\sin(t)) + 7t^2(\cos(t)) = 14t\sin(t) +7t^2\cos(t)$

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Example Question #62 : Spatial Calculus

The acceleration of an object is given by the equation a(t) = 15t^2 + 1 . What is the velocity of the object, if the initial velocity of the object is ?

Possible Answers:

v(t) = 5t^3 +t + 7

v(t) = 7

None of these.

v(t) = 30t

Correct answer:

v(t) = 5t^3 +t + 7

Explanation:

The velocity of the object can be found by integrating the acceleration equation. To integrate this equation we can use the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Therefore the velocity of the object is

$v(t) = \int (15t^2 +1)dt = \frac{15t^{(2+1)}}{2+1} + \frac{1t^{(0+1)}}{0+1} +C = 5t^3 + t +C$ .

The value of can be found using the initial velocity of the object.

v(0) = 5(0)^3 + 0 + C = 0 + 0 + C= 7

Therefore C = 7 and v(t) = 5t^3 +t + 7 .

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Example Question #63 : Spatial Calculus

The position of an object is given by the equation $x(t) = -4\sin^2(5t)$ . What is the velocity of the object at time $t = \frac{\pi }{2}$ ?

Possible Answers:

Correct answer:

Explanation:

The velocity of the object can be found by differentiating the object's position equation. To accurately differentiate the position equation we can use the chain rule and the power rule where if

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x) \times g'(x)$

and where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Therefore the velocity equation of the object is

$v(t) = -4(2)\sin^{(2-1)}(5t)(\cos(5t))(5(1)t^{(1-1)}) = -8\sin(5t)\cos(5t)(5)$ .

We now find the velocity of $t = \frac{\pi}{2}$ ,

$v\left(\frac{\pi}{2}\right) = -8\sin\left(5\left(\frac{\pi}{2}\right)\right)\cos\left(5\left(\frac{\pi}{2}\right)\right)(5)$

$\\= -8\sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi}{2}\right)(5) \\ = -8(1)(0)(5) = 0$

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Example Question #61 : Velocity

The acceleration of an object is given by the equation a(t) = t + t^2 . What is the velocity of the object at t= 1 if the initial velocity of the equation is ?

Possible Answers:

$\frac{5}{6}$

$\frac{4}{3}$

Correct answer:

$\frac{5}{6}$

Explanation:

The velocity of the equation can be found by integrating the acceleration equation. To integrate the acceleration equation, we can use the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Therefore the velocity equation is

$v(t) = \frac{t^{(1+1)}}{1+1} + \frac{t^{(2+1)}}{2+1} +C = \frac{t^2}{2} + \frac{t^3}{3} +C$ .

We can solve for the value of using the initial velocity of the object.

$v(0) = \frac{(0)^2}{2} + \frac{(0)^3}{3} +C = 0 + 0 +C = 0$

Therefore C = 0 and $v(t) = \frac{t^2}{2} +\frac{t^3}{3}$ .

We can now solve velocity at time t =1 by inputting this into the velocity equation.

$v(1) = \frac{(1)^2}{2} + \frac{(1)^3}{3} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$

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Example Question #65 : Calculus

The jerk of an object is given by the equation j(x) = 60t . What is the velocity of the object at time t =2 , if the object has an initial acceleration of and a velocity of at time t = 1 ?

Possible Answers:

265

115

145

Correct answer:

Explanation:

The velocity of the object can be found by integrating the jerk of the object twice. To do this we must use the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Therefore the acceleration of the object is

$a(t) = \int (60t)dt = \frac{60t^{(1+1)}}{1+1} +C = 30t^{2} +C$ .

We can find the value of using the initial acceleration of the object.

a(0) = 30(0)^2 + C = 0 +C = 0

Therefore C = 0 and a(t) = 30t^2 .

Repeating the integration for the acceleration equation gives us

$v(t) = \int (30t^2)dt = \frac{30t^{(2+1)}}{2+1} +C = 10t^3 +C$ .

To find the value of we use the velocity of the object at t =1 .

v(1) = 10(1)^3 +C = 10 +C = 5

Therefore C = -5 and v(t) = 10t^3 -5 .

Now we can find the velocity at t =2 .

v(2) = 10(2)^3 -5 = 75

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Example Question #64 : Calculus

The jerk of an object is given by the equation j(x) = 10 . What is the velocity of the object, if the the acceleration at time t = 3 is and the velocity at time t = 1 is ?

Possible Answers:

v(t) = 5t^2 + 3t -1

v(t) = 5t^2 - 30t +32

v(t) = 5t^2 + 30t -28

v(t) = 0

Correct answer:

v(t) = 5t^2 - 30t +32

Explanation:

To find the velocity we can integrate the jerk of the equation twice. To integrate the equation for jerk we can use the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Therefore the acceleration equation is

$a(t) = \int (10t^0)dt = \frac{10t^{(0+1)}}{0+1} + C = 10t +C$ .

Using the acceleration at time t = 3 , we can find the value of .

a(3) = 10(3) + C =30 + C = 0

Therefore C = -30 and a(t) = 10t -30 .

Repeating this integration for the velocity equation we obtain

$v(t) = \int (10t-30)dt = \frac{10t^{(1+1)}}{1+1} - \frac{30t^{(0+1)}}{0+1} + C = 5t^2 - 30t +C$ .

We can solve for this value of the using the same method as before, but with the velocity at time t =1 .

v(1) = 5(1)^2 - 30(1) +C = 5 - 30 +C = -25 +C = 7

Therefore C = 32 and v(t) = 5t^2 - 30t +32 .

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Example Question #64 : Spatial Calculus

The position of an object is given by the equation $x(t) = \sin^2(t)\cos(t)$ . What is the velocity of the object?

Possible Answers:

$v(t) = 2\sin(t)\cos^2(t) + \sin^3(t)$

$v(t) = -2\sin^2(t)\cos^2(t)$

$v(t) = \sin(t)\cos^2(t) + \sin^3(t)$

$v(t) = 2\sin(t)\cos^2(t) - \sin^3(t)$

Correct answer:

$v(t) = 2\sin(t)\cos^2(t) - \sin^3(t)$

Explanation:

The velocity of the object can be found by differentiating the position equation. To differentiate this equation accurately we can use the power rule and the product rule where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$

and if

$f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$ .

Therefore the velocity of the object is

$\\v(t) = x'(t) = (2)\sin^{(2-1)}(t)(\cos(t))(\cos(t)) + \sin^2(t)(-\sin(t)) \\ \Rightarrow 2\sin(t)\cos^2(t) - \sin^3(t)$

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Example Question #65 : Spatial Calculus

The position of an object is given by the equation $x(t) = \frac{e^{3t}}{2t}$ . What is the velocity of this object?

Possible Answers:

$v(t) = \frac{2te^{3t}-2e^{3t}}{4t^2}$

$v(t) = \frac{3te^{3t}-e^{3t}}{4t^2}$

$v(t) = \frac{6te^{3t}-2e^{3t}}{4t^2}$

$v(t) = \frac{e^{3t}}{2t^2}$

Correct answer:

$v(t) = \frac{6te^{3t}-2e^{3t}}{4t^2}$

Explanation:

The velocity of this object can be found by differentiating the object's position. To accurately differentiate this equation we can use the quotient rule, the chain rule, and the power rule where if

$f(x) = \frac{m(x)}{z(x)} \rightarrow f'(x) = \frac{m'(x)z(x) - m(x)z'(x)}{[z(x)]^2}$ .

Where the chain rule is as follows,

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x) \times g'(x)$ .

The power rule is,

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Therefore the velocity of the object is

$v(t) = x'(t) = \frac{e^{3t}(3(1)t^{(1-1)})(2t) - e^{3t}(2(1)t^{(1-1)})}{(2t)^2} = \frac{6te^{3t}-2e^{3t}}{4t^2}$ .

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Example Question #66 : Spatial Calculus

The position of an object is given by the equation $x(t) = \frac{\cos(e^{2t})}{t}$ . What is the velocity of the object?

Possible Answers:

$v(t) = \frac{-2\sin(e^{2t}) - \cos(e^{2t})}{t^2}$

$v(t) = \frac{2t\sin(e^{2t}) - \cos(e^{2t})}{t^2}$

$v(t) = \frac{-2te^{2t}\sin(e^{2t}) - \cos(e^{2t})}{t^2}$

$v(t) = \frac{2t\sin(e^{2t}) + \cos(e^{2t})}{t^2}$

Correct answer:

$v(t) = \frac{-2te^{2t}\sin(e^{2t}) - \cos(e^{2t})}{t^2}$

Explanation:

The velocity of the object can be found by differentiating the object's position. To accurately differentiate the object's position we can use the quotient rule, the chain rule, and power rule.

The quotient rule is,

$f(x) = \frac{m(x)}{z(x)} \rightarrow f'(x) = \frac{m'(x)z(x) - m(x)z'(x)}{[z(x)]^2}$ .

The chain rule is,

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x) \times g'(x)$ .

The power rule is,

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Also remember that when differentiating

$e^u \rightarrow f'(x)=e^u\cdot \frac{du}{dx}$ .

Therefore the object's velocity equation is

$v(t) = x'(t) = \frac{-\sin(e^{2t})(2e^{2t})(t)-\cos(e^{2t})(1)}{t^2} = \frac{-2te^{2t}\sin(e^{2t}) - \cos(e^{2t})}{t^2}$ .

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #61 : Spatial Calculus

Example Question #2 : Derivatives

Example Question #62 : Spatial Calculus

Example Question #63 : Spatial Calculus

Example Question #61 : Velocity

Example Question #65 : Calculus

Example Question #64 : Calculus

Example Question #64 : Spatial Calculus

Example Question #65 : Spatial Calculus

Example Question #66 : Spatial Calculus

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