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Calculus 1 : Rate of Change

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Example Questions

Example Question #801 : How To Find Rate Of Change

The position of a car is defined by the equation $r(t)=5t^{2}-7t+6$ . What is the instantaneous velocity of the car at t=4 ?

Possible Answers:

$v_{ins}=30$

$v_{ins}=33$

$v_{ins}=0$

$v_{ins}=-4$

Correct answer:

$v_{ins}=33$

Explanation:

The instantaneous velocity of the car is the first derivative of the position at a given point.

In this problem,

$r(t)=5t^{2}-7t+6$

$v_{ins}=r'(t)=10t-7$

$v_{ins}=r'(4)=10*4-7=40-7=33$

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Example Question #801 : How To Find Rate Of Change

Find $\frac{dy}{dx}$ for $f(x)= \frac{ln(x)}{x^8}$ .

Possible Answers:

$\frac{-8}{x^9}+\frac{ln(x)}{x^8}$

$\frac{8ln(x)}{x^9}$

$\frac{x-8ln(x)}{x^9}$

$\frac{1-8ln(x)}{x^9}$

Correct answer:

$\frac{1-8ln(x)}{x^9}$

Explanation:

To solve this problem, we can use either the quotient rule or the product rule. For this solution, we will use the product rule.

The product rule states that $\frac{d(u*v)}{dx}=u'v+v'u$ .

In this case, let $u=x^{-8}\Rightarrow u'=-8x^{-9}$ and $v=ln(x)\Rightarrow v'=\frac{1}{x}$ .

Putting both of these together, we get

$\frac{d(x^{-8}ln(x))}{dx}=-8x^{-9}ln(x)+\frac{1}{x}x^{-8}=\frac{1}{x^{9}}+\frac{-8ln(x)}{x^9}=\frac{1-8ln(x)}{x^9}$ .

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Example Question #801 : How To Find Rate Of Change

Find the slope of the line tangent to the curve of the multivariable function f(x,y) at

the point $(\frac{\sqrt{\pi }}{2},\frac{\sqrt{\pi }}{2})$ .

1=sin(x^2+y^2)

Possible Answers:

-1

-1/2

None of the other answers

1/2

Correct answer:

-1

Explanation:

To find the slope of the tangent line at the specified point, we must first verify that the specified point actually exists on the curve. We check that

$1=sin(x^2+y^2)=sin((\frac{\sqrt{\pi }}{2})^2+(\frac{\sqrt{\pi }}{2})^2)=sin(\frac{\pi }{2})=1$

Since the verification checks out, the problem has a solution and we can continue with Implicit Differentiation.

Recall that for Implicit differentiation, if we have a function in terms of y, we have that it's derivative with respect to x is

$\frac{d}{dx}[f(y)]=\frac{df}{dy}*\frac{dy}{dx}$

Applying this to the given function, we have that

$\frac{d}{dx}[1=sin(x^2+y^2)]$

We must also utilize the Chain Rule to obtain the derivative; we get that

$0=cos(x^2+y^2)*[2x+2y\frac{dy}{dx}]$

Algebraically, we divide the cosine term to begin isolating dy/dx. We then get that

$0=2x+2y\frac{dy}{dx}$

$-2x=2y\frac{dy}{dx}$

$\frac{dy}{dx}=-\frac{x}{y}$

To obtain the slope of the tangent line, we substitute the specified point (x,y) for x and y respectively.

$\frac{dy}{dx}=-\frac{\frac{\sqrt{\pi }}{2}}{\frac{\sqrt{\pi }}{2}}=-1$

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Example Question #804 : How To Find Rate Of Change

A balloon's radius is increasing at a rate of 5 cm/s at the exact moment when the radius of the balloon is 1 cm. Assuming that the balloon is a sphere, at what rate is the volume increasing?

Possible Answers:

$4\pi cm^3/s$

$20\pi cm^2/s$

$4\pi cm^2/s$

$12\pi cm^3/s$

$20\pi cm^3/s$

Correct answer:

$20\pi cm^3/s$

Explanation:

The volume of a sphere is

$V=\frac{4}{3}\pi r^3$

which can also be written as a function with respect to time, i.e..

$V(t)=\frac{4}{3} \pi r(t)^3$ .

If we take the derivative of this, then

$V'(t)=4 \pi r(t)^2 \frac{dr}{dt}$ .

The problem tells us though that the rate of change of the radius is 5 cm/s and $r=1 \text{cm}$ .

Plugging in these values we find that

$V'(t)=4 \pi (1cm)^2\times 5 cm/s= 20\pi cm^3/s$ .

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Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #801 : How To Find Rate Of Change

Example Question #801 : How To Find Rate Of Change

Example Question #801 : How To Find Rate Of Change

Example Question #804 : How To Find Rate Of Change

All Calculus 1 Resources

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