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Chain Rule, Log Differentiation & Rate of Change · Question 21 of 130

What is the particle's velocity ds/dt?

Question 21

MediumOriginal question

A particle's distance from a point after tt seconds is s=t2+16s = \sqrt{t^{2} + 16} metres.

What is the particle's velocity dsdt\frac{ds}{dt}?

Options for question 21

Show the answer and explanation
CorrectNot quiteAnswer A

Correct answer: A

Explanation

  1. ddtu=u′2u\frac{d}{dt}\sqrt{u} = \frac{u'}{2\sqrt{u}} with u=t2+16u = t^{2} + 16.

  2. 2t2t2+16=tt2+16\frac{2t}{2\sqrt{t^{2} + 16}} = \frac{t}{\sqrt{t^{2} + 16}}.

Worth noting

12t2+16\frac{1}{2\sqrt{t^{2} + 16}} leaves out the inner 2t2t.

Concept
The chain rule
Typical time
120 s
Your time
—
The rule this testsThe chain rule
Practise this chapter →

Ten questions from Chain Rule, Log Differentiation & Rate of Change, at your own pace.

Practise this chapter →

Ten questions from Chain Rule, Log Differentiation & Rate of Change, at your own pace.